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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
Find the slope of the line passing through the points `(i) (-2,3) and (8,-5)" "(ii) (4,-3) and (6,-3)" "(ii) (3,-1) and (3,2)` |
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Answer» (i) Let `A(-2,3) and B(8,-5)` be tha given points. Then, slope of AB=`(-5-3)/(8-(-2))=(-8)/(10)=(-4)/(5)" "[thereforem=((y_(2)-y_(1)))/((x_(2)-x_(1)))]` (ii) Let C(4,-3) and D(6,-3) be the given points. Then, slope of CD=`(-3-(-3))/(6-4)=(-3+3)/(2)=(0)/(2)=0` ALTER: The points C(4,-3) and D(6,-3) have the same y-coordinate. So, CD is a line parallel to the line x-axis. Hence, it slope is 0. (iii) Let P(3,-1) and Q(3,2) , which is not defined. Slope of PQ=`=(2-(-1))/(3-3)=(3)/(0),` which is not defined ALTER: The points P(3,-1) and Q(3,2) have the same x-coordinates. So, PQ is a vertical line. Hence, its slope is not defined. |
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| 152. |
The line `(k + 1)x + ky-2k^2-2=0` passes through a point regardless of the value k. Which of the following is the line with slope 2 passing through the point? |
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Answer» the line given is `(k+1)^2x +ky-2k^2 -2 = 0` as `y=mx+c` `ky= -(k+1)^2x+ (2k^2+2)` `y= -(k+1)^2x/k + (2k^2+2)/k` where `m= -(k+1)^2/k = 2` `-(k+1)^2 = 2k` `-(k^2+2k+1) = 2k` `-k^2 -2k-2k-2=0` `k^2 +4k+1= 0` `k= (-4+-sqrt(16-4*1))/2` `k= -2 +- sqrt 3` c is given `c=(2k^2 +2)/k= 2(k+1/k)` for `k=-2 +sqrt3` `c = 2(-2 + sqrt3 + (1/(-2+sqrt3))*(-2-sqrt3)/(-2 - sqrt3)` `=2[-2 + sqrt3 + (-2 - sqrt3)]` `= 2(-4) = -8` for `k=-2 -sqrt3` `c = 2(-2 - sqrt3 + (1/(-2-sqrt3))*(-2-+qrt3)/(-2 + sqrt3)` `=2[-2 - sqrt3 + (-2 + sqrt3)]` `= 2(-4) = -8`equation of line is `y= 2x-8` option A is correct |
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| 153. |
Find the equation of the straight line which passes through the originand makes angle `60^0`with the line `x+sqrt(3)y+sqrt(3)=0`. |
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Answer» Correct Answer - x=0, x`-sqrt(3)` y =0 The given line is `x+sqrt(3)y+3sqrt(3) = 0.` `"or " y=(-(1)/(sqrt(3)))x-3` Therefore, the slope of (1) is `-1//sqrt(3).` Let the slope of the required line be m. Also, the angle between these line is `60^(@)`. Therefore, `" tan " 60^(@) = |(m-(-1//sqrt(3)))/(1+m(1//sqrt(3)))|` `"or " sqrt(3) = |(sqrt(3)m+1)/(sqrt(3)-m)|` `"If "(sqrt(3)m+1)/(sqrt(3)-m) =sqrt(3)` `"or "m=(1)/(sqrt(3))` Using y=mx +c, the equation of the required line is `y=(1)/(sqrt(3))x +0` `"i.e.," x-sqrt(3) y = 0` (as the line passes through the origin, c=0) `(sqrt(3)m-1)/(sqrt(3)-m) = -sqrt(3)` `"or " sqrt(3)m+1 =-3+sqrt(3)m` Therefore, finite m does not exist. Therefore, the slope of the required line is infinity. Thus, the required line is a vertical line. This line passes through the origin. Therefore, the equation of the required line is x=0. |
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| 154. |
Find the equation of a line passing through the point (4,-3) and having slope 2. |
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Answer» We knot that the equation of a line with slope m and passsing through the point `(x_(1),y_(1))` is given by `(y-y_(1))=m(x-x_(1))` Hence, m=2,`x_(1)=4 and y_(1)=3` Hence, the required equation is `(y-3)=2(x-4), i.e. 2x-y-5=0` |
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| 155. |
Find the equation of a line which makes an angle of `135^(@)` with the x-axis and passes through the point (3,5). |
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Answer» Here, `m=tan 135^(@)=tan(180^(@)-45^(@))=-tan 45^(@)=-1` Hence, the required equation is `(y-5)/(x-3)=-1 Leftrightarrow (y-5)=3-x Leftrightarrow x+y-8=0` |
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| 156. |
Find the slope of the line whose inclination is π/4 |
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Answer» Given, inclination (0) = π/4 ∴ Slope of the line = tan θ = tan π/4 =1 |
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| 157. |
Line through the points (-2,6) and (4,80 is perpendicular to the linethrough the points (8,12) and `(x , 24)dot`Find the value of `xdot` |
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Answer» Let A(-2,6), B(4,8), C(8,12) and D(x,24) be the given points. Let `m_(1) and m_(2)` be the slopes of AB and CD respectively. Then, `m_(1)="slope of AB"=((8-6))/(4-(-2))=(2)/(6)=(1)/(3)` `m_(2)="slope of CD"=((24-12))/((x-8))=(12)/((x-8))` `"Now", AB CD Leftrightarrow m_(1)m_(2)=-1` `Leftrightarrow (1)/(3)xx(12)/((x-8))=-1` `Leftrightarrow -x+8=4 Leftrightarrow x=4` Hence, the required value of x is 4. |
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| 158. |
Show that the joining (2,-5) and (-2, 5) is perpendicular to the linejoining (6,3) and (1,1). |
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Answer» Let A(2,-5), B(-2,5),C(6,3) and D(1,1) be the given points. let `m_(1) and m_(2)` bet the slopes of AB and CD respectively. Then, `m_(1)="slope of AB"=(5-(-5))/(-2-2)=(10)/(4)=-(5)/(2)` `m_(2)="slope of CD"=(1-3)/(1-6)=(-2)/(-5)=(2)/(5)` `therefore m_(1)m_(2)=((-5)/(2))xx(2)/(3)=-1` Hence, `AB"perpendicular"CD` |
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| 159. |
Find the angle between the X-axis and the line joining the points `(3,-1)a n d (4,-2)dot` |
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Answer» Let A(3,-1) and B(4,-2) be the given points and let m be the slope of the line AB. Then, `m=(-2-(-1))/(4-3)=(-2+1)/(1)=-1" "[therefore m=((y_(2)-y_(1)))/(x_(2)-x_(1))]` Let `theta` be the angle between the x-axis and the line AB. Then, `tan theta=m=-1=tan 45^(@)=tan(180^(@)-45^(@))=tan135^(@)` `therefore theta=135^(@)` Hence, teh required angle is `135^(@)` |
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| 160. |
Show that the line joining (2,-3) and (-5,1) is parallel to the linejoining (7,-1) and (0,3). |
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Answer» Let A(2,-3), B(-5,1), C(7,-1) and D(0,3) be the given points. Then, slope of AB=`(1-(-3))/(-5-2)=(4)/(-7)=-(4)/(7)` And, slope of CD=`(3-(-1))/(0-7)=(4)/(-7)=-(4)/(7)` `therefore` slope of AB=slope of CD. Hence, `AB ||CD` |
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| 161. |
Find the equation of a line which is equidistant from the lines x=-2 and x=6. |
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Answer» Correct Answer - x=2 Midpoint of `A(-2,0) and B(6,0) is M((-2+6)/(2),0)` i.e. M(2,0), So, the required line is x=2. |
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| 162. |
Find the equation of the line passing through the point P(4, -5) and parallel to the line joining the points A(3, 7) and B(-2, 4). |
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Answer» Correct Answer - 3x-5y-37=0 Slope of the given line=slope of AB=`(4-7)/(-2-3)=(3)/(5)`, Thus, m`=(3)/(5)` `therefore` required equation is `(y+5)/(x-4)=(3)/(5) Leftrightarrow 3x-5y-37=0`. |
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| 163. |
Find the equation of a line which is equidistant from the lines y=8 and y=-2 |
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Answer» Correct Answer - y=3 Midpoint of A(0,8) and B(0,-2) is M `(0,(8-2)/(2))`, i.e M(0,3) |
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| 164. |
Find the equation of a line which cuts off intercept 5 on the x-axis and makes an angle of `60^(@)` with the positive direction of the x-axis. |
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Answer» Correct Answer - `sqrt3x-y-5sqrt3=0` The given line passes through the point (5,0) and `m=tan 60^(@)=sqrt3`. So, the required equation is `(y-0)/(x-5)=sqrt3 Leftrightarrow y=sqrt3(x-5)`. |
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| 165. |
Find the equation of the line whose inclination is `(5pi)/(6)` and which makes an intercept of 6 units on the negative direction of the y-axis. |
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Answer» Correct Answer - `x+sqrt3y+6sqrt3=0` `m="tan "(5pi)/(6)=tan(pi-(pi)/(6))=-"tan"(pi)/(6)=-(1)/(sqrt3).And,c=-6` |
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| 166. |
The algebraic sum of the perpendicular distances from `A(x_1, y_1)`, `B(x_2, y_2)` and `C(x_3, y_3)` to a variable line is zero. Then the line passes through (A) the orthocentre of `triangleABC` (B) centroid of `triangleABC` (C) incentre of `triangleABC` (D) circumcentre of `triangleABC`A. the orthocentre of `Delta ABC`B. the centroid of `Delta ABC `C. the circumcentre of `DeltaABC`D. none of these |
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Answer» Correct Answer - B Let `Px + Qy = 1` be a variable line, where P,Q are variable. By hypothesis , we have . `(Px_(1)+Qy_(1)-1)/(sqrt(p^(2) +Q^(2)) +(p^(2)+Q^(2)))+(Px_(3)+Qy_(3)-1)/(sqrt(P^(2)+Q^(2)))=0` `rArr P(x_(1) + x_(2) + x_(3)) +Q(y_(1) + y_(2)+y_(3)) = 3` `rArr p((x_(1)+x_(2)+x_(3))/(3))+Q((y_(1)+y_(2)+y_(3))/(3))=1` `rArr ((x_(1) + x_(2)+x_(3))/(3),(y_(1)+y_(2)+y_(3))/(3))` lies on `Px + Qy=1` Hence, the variable line passig thorought the the centriod of the triangle ABC. |
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| 167. |
If `a , b ,c`are in harmonic progression, then the straight line `((x/a))_(y/b)+(l/c)=0`always passes through a fixed point. Find that point. |
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Answer» a,b,c are in HP. Then, `(2)/(b) = (1)/(a) + (1)/(c)" "(1)` The given line is `(x)/(a) + (y)/(b) + (1)/(c) = 0" "(2)` Subtracting (1) from (2), we get `(1)/(a)(x-1) + (1)/(b)(y+2) =0` Since `a ne 0 " and " b ne 0`, we get x-1 =0 and y+2 = 20 or x =1 and y=-2 |
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| 168. |
A variable line passes through a fixed point P. The algebraic sum of the perpendiculars drawn from the points (2,0), (0,2) and (1,1) on the line is zero. Find the coordinate of the point P. |
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Answer» Let the equation of the variable line be ax+by+c = 0 Then according to the given condition, we get `(2a+c)/(sqrt(a^(2) + b^(2))) + (2b+c)/(sqrt(a^(2) + b^(2))) + (-2a-2b+c)/(sqrt(a^(2)+b^(2))) = 0` or c=0 which shows that the line passes through (0,0) for all values of a and b. |
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| 169. |
The equation of lines on which the perpendiculars from the origin make `30^0`angle with the x-axis and which form a triangle of area `(50)/(sqrt(3))`with the axes are`sqrt(3)x+y-10=0``sqrt(3)x+y+10=0``x+sqrt(3)y-10=0`(d) `x-sqrt(3)y-10=0`A. `x + sqrt3 y pm 10 = 0`B. `sqrt3 x + y pm 10 = 0`C. `x pm sqrt3 y - 10 = 0`D. none of these |
| Answer» Correct Answer - B | |
| 170. |
If three lines whose equations are `y=m_1x+c_1,y=m_2x+c_2`and `y=m_3x+c_3`are concurrent, then show that `m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0`. |
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Answer» The given line are: `m_(1)x-y+c_(1)=0......(i)` `m_(2)x-y+c_(2)=0......(ii)` `m_(3)x-y+c_(3)=0......(iii)` On solving (i) and (ii) by cross multiplication, we get `(x)/((-c_(2)+c_(1)))=(y)/((m_(2)c_(1)-m_(1)c_(2)))=(1)/((-m_(1)+m_(2)))` `Rightarrow x=((c_(1)-c_(2)))/((m_(2)-m_(1)))and y=((m_(2)c_(1)-m_(1)c_(2)))/((m_(2)-m_(1)))` Thus, the point of intersection of (i) and (ii) is P `((c_(1)-c_(2))/(m_(2)-m_(1)),(m_(2)c_(1)-m_(1)c_(2))/(m_(2)-m_(1)))` Since the given three lines meet at a point, the point P must therefore lie on (ii) also. `therefore m_(3).((c_(1)-c_(2)))/((m_(2)-m_(1)))-((m_(2)c_(1)-m_(1)c_(2)))/((m_(2)-m_(1)))+c_(3)=0` `Rightarrow m_(3)(c_(1)-c_(2))-(m_(2)c_(1)-m_(1)c_(2))+c_(3)(m_(2)-m_(1))=0` `Rightarrow m_(1)(c_(2)-c_(3))-m_(2)(c_(3)-c_(1))-m_(3)(c_(1)-c_(2))=0` |
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| 171. |
Consider the system of equations `a_(1) x + b_(1) y + c_(1) z = 0` `a_(2) x + b_(2) y + c_(2) z = 0` `a_(3) x + b_(3) y + c_(3) z = 0` If `|(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3))| =0`, then the system hasA. Statement -1 is True , Statement - 2 is true , Statement- 2 is a correct explanation for statement - 4B. Statement-1 is True , Statement-2 is True , Statement -2 is not a correct explanation for Statement - 1 .C. Statement-1 is True , Statement - 2 is False .D. Statement - 1 is False , Statement -2 is True . |
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Answer» Correct Answer - A The area of triangle formed by the lines `a_(1)x + b_(1) y + c_(1) = 0 , a_(2) x + b_(2)y + c_(2) = 0` and `a_(3) x + b_(3) y + c_(3) = 0` is `Delta = (1)/(2C_(1) C_(2)C_(3)) |{:(a_(1), b_(1), c_(1)), (a_(2) ,b_(2), c_(2)) , (a_(3), b_(3), c_(3)):}|^(2)` where `C_(1) , C_(2) , C_(3)` are cofactors of `c_(1) , c_(2) , c_(3)` respectively in the matrix `[{:(a_(1), b_(1), c_(1)), (a_(2) ,b_(2), c_(2)) , (a_(3), b_(3), c_(3)):}]` If the lines are concurrent , then `Delta = 0 implies |{:(a_(1), b_(1), c_(1)), (a_(2) ,b_(2), c_(2)) , (a_(3), b_(3), c_(3)):}|` = 0 So , statement -2 is true . Also , statement - 1 is true and statement - 2 is a correct explanation for statement -1 . |
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| 172. |
Pis a point inside the triangle ABC. Lines are drawn through P, parallel to the sides of the triangle.The three resulting triangles with the vertex at P have areas `4,9 and 49 sq`. units. The area of thetriangle ABC is -A. `2sqrt(3)`B. 12C. 24D. 144 |
| Answer» Correct Answer - D | |
| 173. |
Show that the lines `a_(1)x+b_(1)y+c_(1)=0 and a_(2)x+b_(2)y+c_(2)=0,"where" b_(1),b_(2) ne 0 "are (i) parallel, if"(a_(1))/(b_(1))=(a_(2))/(b_(2))" (ii) perpendicular, if "a_(1)a_(2)+b_(1)b_(2)=0` |
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Answer» Let the slopes of the given lines be `m_(1) and m_(2)` respectively. Now, `a_(1)x+b_(1)y+c_(1)=0 Rightarrow y=(-a_(1))/(b_(1)).x-(c_(1))/(b_(1))` `and a_(2)x+b_(2)y+c_(2)=0 Rightarrow y=(-a_(2))/(b_(2)).x-(c_(2))/(b_(2))` `therefore m_(1)=(-a_(1))/(b_(1)) and m_(2)=(-a_(2))/(b_(2))` (i) Given lines are parallel, if `m_(1)=m_(2)` which given `(-a_(1))/(b_(1))=(-a_(2))/(b_(2)) or (a_(1))/(a_(2))=(b_(1))/(b_(2))` (ii) Given line are perpendicular, `if m_(1)m_(2)=-1` which gives `((-a_(1))/(b_(1))).((-a_(2))/(b_(2)))=-1 or a_(1)a_(2)+b_(1)b_(2)=0` |
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| 174. |
Find the equations of the line which passes through the point `(3,4)`and the sum of its intercepts on the axes is`14`. |
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Answer» Let the intercepts made by the line on the x-axis and y-axis be a and (14-a) respectively. Then, its equation is `(x)/(a)+(y)/((14-a))=1" "["intercept form"]` Since it passes through the point (3,4), we have `(3)/(a)+(4)/((14-a))=1 Leftrightarrow 3(14-a)+4a=a(14-a)` `Leftrightarrow a^(2)-13a+42=0 Leftrightarrow (a-6)(a-7)=0` `Leftrightarrow a=6 or a=7` `"Now," a=b Leftrightarrow 14-6=8` `"And" a=7 Leftrightarrow 14-7=7` So, the required equation is `(x)/(6)+(y)/(8)=1 or (x)/(7)+(y)/(7)=1` `i.e. 4x+3y-24=0 or x+y-7=0` |
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| 175. |
Find the equation of the line passing through the point (2,-5) and parallel to the line 2x-3y=7. |
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Answer» We have `2x-3y=7 Rightarrow y=(2)/(3)x=(7)/(3)` `therefore` slope of the given line=`(2)/(3)` and the slope of a line parallel to the given line=`(2)/(3)` . Now, the equation of the line slope `(2)/(3)` and passing through the points (2,-5) is given by `(y-(-5))/(x-2)=(2)/(3) Rightarrow (y+5)/(x-2)=(2)/(3)" "[therefore ((y-y_(1)))/((x-x_(1)))=m]` `Rightarrow 3(y+5)=2(x-2)` `Rightarrow 2x-3y-19=0`, which is the required equation. |
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| 176. |
Find the equationof the line, which makes intercepts `3`and 2 on thex and yaxes respectively. |
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Answer» We know that the equation of a line making intercepts a and b on the x-axis and y-axis respectively, is `(x)/(a)+(y)/(b)=1` Here a=2 and b=-3 Hence, the required equation is `(x)/(2)+(y)/(-3)=1 Leftrightarrow (x)/(2)-(y)/(3)=1 Leftrightarrow 3x-2y-6=0` |
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| 177. |
The equations of the lines through `(-1,-1)` and making angle `45^@` with the line `x+y=0` are given byA. `x^(2) -xy + x - y =0`B. `xy -y^(2) + x - y = 0`C. `xy + xy+ y=0`D. `xy + x+ y +1=0` |
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Answer» Correct Answer - D The equations of the lies passing through `(x_(1),y_(1))` and making an angle `alpha` with y = mx + c are given by `y-y_(1) =(m pm tan alpha)/(1 barpm m tan alpha )(x-x_(1))` So, the eqauations of the required lines are `y+ 1 = 0` and x + 1=0 The combined equation of these tow lines is `(x +1) (y+1) = 0` i.e., `xy + x+y +1=0` |
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| 178. |
Find the slope of the lines whose iclination is given : `(i) 45^(@)` `(ii) 60^(@)` `(iii) 120^(@)` |
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Answer» Let m be the slope of the line. Then, (i) `m=tan 45^(@)=1` `(ii)=m=tan 60^(@)=sqrt3` `(iii)=m=tan 150^(@)=tan(180^(@)-30^(@))=-tan 30^(@)=(-1)/(sqrt3)` |
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| 179. |
Find the tion of the line passing through the point `(-2,-4)`and perpendicular to the line `3x-y+5=0` |
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Answer» We have lt`3x-y+5=0 Rightarrow y=3x+5` `therefore` slope of the given line=3 `[therefore y=mx+c]` and slope of the line perpendicular to the given line`=-(1)/(3) " "[therefore m_(2)=-(1)/(m_(1))]` Now, the equation of a line with slope `-(1)/(3)` and passing through the point (-2,-4) is given by `(y-(-4))/(x-(-2))=-(1)/(3) Rightarrow (y+4)/(x+2)=(-1)/(3)" "[therefore ((y-y_(1)))/((x-x_(1)))=m]` `Rightarrow 3(y+4)=-(x+2)` `Rightarrow x+3y=14=0`, which is the required equation. |
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| 180. |
Find the equation of the line y-intercept is -3 and which is perpendicular to the line `3x-2y+5=0` |
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Answer» `3x-2y+5=0 Rightarrow y=(3)/(2)x+(5)/(2)` `therefore` slope of the given time=`(3)/(2)` `therefore` slope of a line perpendicular to the given line `=(-2)/(3)" "[therefore m_(2)=(-1)/(m_(1))]` Now, the equatoin of a line with slope `(-2)/(3)` and y-intercept -3 is given by `y=(-2)/(3)x-3" "[therefore y=mx+c]` `Rightarrow 2x+3y+9=0`, which is the required equation. |
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| 181. |
Find the equation of the lines through the point (3, 2) which make an angle of `45^@`with the line `x-2y=3`. |
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Answer» Let the slope of the required line be m, Then, its equation is `(y-2)/(x-3)=m.......(i)` Given line is `x-2y=3 Rightarrow y=(1)/(2)x-(3)/(2).....(ii)` Clearly, the slope of this line is `(1)/(2)` It is given that the angle between (i) and (ii) is `45^(@)` `therefore |(m-(1)/(2))/(1+(1)/(2)m)|=tan 45^(@)" "["using tan "theta=|(m_(2)-m_(1))/(1+m_(1)m_(2))|]` `Leftrightarrow (2m-1)/(2+m)=1 or (2m-1)/(2+m)=-1 Leftrightarrow m=3 or m=(-1)/(3)` `therefore ` the required equation is `(y-2)/(y-3)=3 or (y-2)/(x-3)=(-1)/(3)` `i.e. 3x-y-7=0 or 3y-9=0` |
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| 182. |
Find equation of the line perpendicular to the line `x" "" "7y" "+" "5" "=" "0`and having x intercept 3. |
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Answer» `x-7y+5=0 Rightarrow y=(1)/(7)x+(5)/(7)` `therefore` slope of the given line=`(1)/(7)` `therefore` slope of a line perpendicular to the given line=`-7[therefore m_(2)=(-1)/(m_(1))]` Thus, m=-7 and d=3, where d=x, intercept. Now, the equation of a line with slope -7 and x-intercept 3 is given by `y=-7(x-3)" "[therefore y=m(x-d)]` `Rightarrow 7x+y-21=0`, where is the required equations. |
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| 183. |
The ends of the base of an isosceles triangle are at `(2a, 0)and (0, a).` The equation of one side is `x = 2a. `The equation of the other side, isA. x + 2y - a = 0B. x + 2y = 2aC. 3x + 4y - 4a = 0D. 3x - 4y + 4a = 0 |
| Answer» Correct Answer - D | |
| 184. |
If the extremities of the base of an isosceles triangle are the points `(2a ,0)`and (0, a), and the equation of one of the side is `x=2a ,`then the area of the triangle is`5a^2s qdotu n i t s`(b) `(5a^2)/2s qdotu n i t s``(25 a^2)/2s qdotu n i t s`(d) none oftheseA. `5a^(2) "sq. units"`B. `5a^(2)//2 "sq. units"`C. `25a^(2)//2 "sq. units"`D. none of these |
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Answer» Correct Answer - B Given vertices are A(2a,0) and B(0,a) Let the coordinates of the third vertex C be (2a,t). Now, AC = BC. Hence, `t = sqrt(4a^(2) + (a-t)^(2)) " or " t =(5a)/(2)` So, the coordinates of the third vertex C are (2a,5a/2). Therefore, area of the triangle is `(1)/(2)||{:(2a, 5a//2, 1),(2a, 0, 1),(0, a, 1):}|| = (5a^(2))/(2)` sq. units |
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| 185. |
A variable line `x/a + y/b = 1` moves in such a way that the harmonic mean of a and b is 8. Then the least area of triangle made by the line with the coordinate axes is(1) 8 sq. unit(2) 16 sq. unit(3) 32 sq. unit(4) 64 sq. unitA. 8 sq. unitB. 16 sq. unitC. 32 sq. unitD. 64 sq. unit |
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Answer» Correct Answer - C Given line is `(x)/(a)+(y)/(b)=1` It meets the axes at point A(a,0) and B(0,b). Area of triangle AOB, `Delta = (1)/(2)` ab Given that 8 is H.M. of a and b. `therefore (1)/(a)+(1)/(b)=(1)/(4)` `rArr b = (4a)/(a-4)` `therefore Delta = (2a^(2))/(a-4)` Differentiating w.r.t. a, we get ` (dDelta)/(da) = 2(2a(a-4)-a^(2))/((a-4)^(2))` `rArr (dDelta)/(da) = (2a(a-8))/((a-4)^(2))` `"If " (dDelta)/(da) = 0, "then " a =8` This is the value of a for which area is minimum `therefore Delta_(min.) = 32 " sq.units."` |
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| 186. |
If the slope of one line of the pair of lines represented by `ax^2 +10xy+ y^2 =0` is four times the slope of the other line, then `a=` |
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Answer» Let `m` is the slope of one line,then the slope of another line will be `4m`. So, `(y-mx)(y-4mx) = ax^2+10xy+y^2` `y^2-4mxy-mxy+4m^2x^2 = ax^2+10xy+y^2` `=>4m^2x^2-5mxy+y^2 = ax^2+10xy+y^2` So, `-5m = 10 ` and `a = 4m^2` `m = -2` and `a= 4**(-2)^2 => a= 16` |
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| 187. |
If the angle between two lines is `pi/4`and slope of one of the lines is `1/2`, find the slope of the other line. |
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Answer» Let `m_1` and `m_2` are the slopes of these two lines and `theta` is the angle between these two lines. Then, `tan theta = |(m_1-m_2)/(1+m_1m_2)|` Here, `theta = pi/4 and m_1 = 1/2` `:. tan(pi/4) = |(1/2-m_2)/(1+1/2m_2)|` `=> 1 = |(1/2-m_2)/(1+1/2m_2)|` `=> (1/2-m_2)/(1+1/2m_2) = 1 or (1/2-m_2)/(1+1/2m_2) = -1` `=>(1/2-m_2) = (1+1/2m_2) or (1/2-m_2) = -(1+1/2m_2)` `=>3/2m_2 = -1/2 or -1/2m_2 = -3/2` `=>m_2 = -1/3 or m_2 = 3` So, slope of other line will be `-1/3` or `3`. |
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| 188. |
Statement -1 : The line 3x + 2y = 24 meets the coordinates axes at A and B , and the perpendicular bisector of AB meets the line through (0,-1) parallel to the x-axis at C . The area of `Delta` ABC is 91 square units . Statement-2 : Area of the triangle with vertices at (a,0) , (0,b) and (a , b) is ab/2 sq. units .A. Statement -1 is True , Statement - 2 is true , Statement- 2 is a correct explanation for statement - 8B. Statement-1 is True , Statement-2 is True , Statement -2 is not a correct explanation for Statement - 1 .C. Statement-1 is True , Statement - 2 is False .D. Statement - 1 is False , Statement -2 is True . |
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Answer» Correct Answer - B The line 3x + 2y = 24 meets the coordinate axes at A (8,0) , B( 0 , 12) . The coordinates of the mid-point D of AB are (4,6) . Equation of the perpendicular bisector of AB is 2x - 3y + 10 = 0 This meets the line through (0 ,-1) parallel to the x-axis i.e. y = -1 at C (-13/2 , -1) . `therefore` Area of `DeltaABC = (1)/(2) AB xx CD` `implies` Area of `Delta ABC = (1)/(2) sqrt(64 + 144) xx sqrt(((13)/(2) + 4)^(2) + (1-6)^(2))` `implies` Area of `Delta ABC = (1)/(2) xx sqrt(208) xx 7 sqrt((9)/(4) + 1) = 91` sq. units So , statement-1 is true . The area of the triangle with vertices at (a,0) , (0, b) and (a , b) is Absolute value of `(1)/(2) |{:(a , b , 1), (0 , b , 1) , (a , b , 1):}| = (1)/(2)` ab sq. units . So , statement - 2 is true . But , statement - 2 is not a correct explanation for statement - 1 . |
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| 189. |
Find the angle between the lines whose slope are `(1)/(2)` and 3. |
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Answer» Let `theta` be the angle between the given lines. `"Let" m_(1)=(1)/(2) and m_(2)=3."Then",` `tan theta=|(m_(2)-m_(1))/(1+m_(1)m_(2))|=|(3-(1)/(2))/(1+3.(1)/(2))|=|((5//2))/((5//2))|=|1|=1` `Rightarrow theta=45^(@)` Hence, the angle between the given lines is `45^(@)` |
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| 190. |
Write the coordinates of the orthocentre of thetriangle formed by points (8,0), (4,6) and (0,0)A. (4,8/3)B. (3,4)C. (4,3)D. (-3,4) |
| Answer» Correct Answer - A | |
| 191. |
Find the area of the quadrilateral whose vertices are A(-4,5), B(0,7), C(5,-5) and D(-4,-2). |
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Answer» Correct Answer - 60.5 sq units Area of quad. ABCD=are`(triangle ABC)+"area"(triangle ACD)` |
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| 192. |
The line `3x+2y=24`meets the y-axis at `A`and the x-axis at `Bdot`The perpendicular bisector of `A B`meets the line through `(0,-1)`parallel to the x-axis at `Cdot`If the area of triangle `A B C`is `A`, then the value of `A/(13)`is________ |
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Answer» Correct Answer - 91 Lines 3x+2y = 24 meets the axes at B(8,0) and A(0, 12). The midpoint of AB is D(4, 6). The equation of perpendicular bisector of AB is `2x-3y+10=0 " " (1)` Now, the line through (0, -1) and parallel to the x-axis is y=-1. The coordinates of C where line (1) meets y =-1 are (-13/2, -1). Now, the area of triangle ABC, `Delta = (1)/(2)||{:(0, 12, 1),(8, 0, 1),(-13//2, -1, 1):}||` `= (1)/(2)|[0-12(8+(13)/(2))+1(-8)]| = (1)/(2) |[-6(29)-8]| = 91` |
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| 193. |
Find the point of intersection of the following pairs of lines: `b x+a y=a b a n d b x+b y=a bdot`A. `x=y+4`B. `(lx+my)(a+b)=(l+m)ab`C. `(x+y)(a+b)=2ab+2`D. `(lx-my)(a-b)=(l-m)ab` |
| Answer» Correct Answer - B | |
| 194. |
The straight lines `x+y-4=0, 3x+y-4=0` and `x+3y-4=0` form a triangle, which isA. isoscelesB. right angledC. equilateralD. none of these |
| Answer» Correct Answer - A | |
| 195. |
A straight line is drawn through the centre of the circle `x^2 + y^2-2ax=0` line `x + 2y = 0 ` parallel to the straight and intersecting the circle at `A` and `B`. Then the area of triangle `AOB` is |
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Answer» `x^2+y^2-2ax=0` `x+2y=0` equation of line x+2y++c=0 a+0+c=0 c=-a We get, x+2y-a=0 From diagram `h=|(-a/sqrt5)` `h=a/sqrt5` b=2a area of triangle=`1/2*B*H` =`1/2*a/sqrt5*2a` =`a^2/sqrt5` |
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| 196. |
The vertices of a `triangleOBC` are `O(0,0) , B(-3,-1), C(-1,-3)`. Find the equation of the line parallel to BC and intersecting the sides OB and OC and whose perpendicular distance from the origin is `1/2`.A. ` x + y + (1)/(2) = 0`B. `x + y - (1)/(2) = 0`C. `x + y - (1)/(sqrt2) = 0`D. `x + y + (1)/(sqrt2) = 0` |
| Answer» Correct Answer - D | |
| 197. |
The vertices of a `triangleOBC` are `O(0,0) , B(-3,-1), C(-1,-3)`. Find the equation of the line parallel to BC and intersecting the sides OB and OC and whose perpendicular distance from the origin is `1/2`. |
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Answer» Equation of line BC `y+1=(-3+1)/(-1+3)(x+3)` `y+1=-x-3` `y=-x-4` `y=-x+c-(1)` `y+x-c=0` `|(-C)/sqrt2|=1/2` `C/sqrt2=1/2` `C=1/sqrt2` `y=-x+1/sqrt2` `sqrt2y=-sqrt2x+1` `sqrt2x+sqrt2y=1`. |
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| 198. |
Given `A-=(1,1)`and `A B`is any line through it cutting the x-axis at `Bdot`If `A C`is perpendicular to `A B`and meets the y-axis in `C`, then the equation of the locus of midpoint `P`of `B C`is`x+y=1`(b) `x+y=2``x+y=2x y`(d) `2x+2y=1`A. `x+y=1`B. `x+y=2`C. `x+y=2xy`D. `2x+2y=1` |
| Answer» Correct Answer - C | |
| 199. |
A is a point on the x-axis with abscissa -8 and B is a point on the y-axis with ordinate 15. Find the distance AB. |
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Answer» Correct Answer - 17 units Given points are A(-8,0) and B(0,15) |
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| 200. |
Statement 1: Each point on the line `y-x+12=0`is equidistant from the lines `4y+3x-12=0,3y+4x-24=0`Statement 2: The locus of a point which is equidistantfrom two given lines is the angular bisector of the two lines.A. Statement -1 is True , Statement - 2 is true , Statement- 2 is a correct explanation for statement - 6B. Statement-1 is True , Statement-2 is True , Statement -2 is not a correct explanation for Statement - 1 .C. Statement-1 is True , Statement - 2 is False .D. Statement - 1 is False , Statement -2 is True . |
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Answer» Correct Answer - A Statement - 2 is true and it is the correct explanation for statement-1 as y - x + 12 = 0 is one of the angle bisectors of the given lines . So , statement -1 is also true . |
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