InterviewSolution
Saved Bookmarks
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.
| 101. |
Show that the points A(6, 1), B(8, 2), C(9, 4) and D(7, 3) are the vertices of a rhombus. Find its area. |
| Answer» Correct Answer - 3 sq units | |
| 102. |
(ii)In what ratio does the x axis divide the line segment joining the points `(2;-3) and (5;6)`. Find the coordinate of a point of intersection |
|
Answer» let the point on x axis be (x,0) the ratio be m:1 using section formula,=>`(6m-3)/(m+n)=0` `m=1/2` the ratio=>`1:2` |
|
| 103. |
The points `(-2,5), (3, -4) and (7, 10)` are the vertices of the triangle then triangle is: |
|
Answer» Let `A(-2,5), B(3,-4) and C(7,10)` are the vertices of the given triangle. Then, `AB = sqrt((3-(-2))^2+(-4-5)^2) = sqrt106` `BC= sqrt((10 - (-4))^2+(7-3)^2) = sqrt212` `AC = sqrt((7-(-2))^2+(10-5)^2) = sqrt106` Now, `AB = AC` Also, `AB^2+AC^2 = BC^2` So, `ABC` is a right angled isoceles triangle. |
|
| 104. |
Using analytical geometry, prove that the diagonals of a rhombus areperpendicular to each other. |
|
Answer» we need to prove ABCD is a rhombus where AB=BC=BD=DA A(0,0) B(S,0) C(A+S,B) D(A,B) `AD=SQRT(a^2+b^2)=s` `a^2+b^2=s^2` AB=`sqrt(s^2+0^2)=s` BC=`sqrt((a+s-s)^2+(b-0)^2)=sqrt(a^2+b^2)=S` CD=`sqrt((a+S-a)^2+(b-b)^2)=sqrt(s^2)=S` DA=`sqrt(a^2+b^2)=sqrt(s^2)=S` secondly we need to prove they are perpendicular `M_(AC)=(b-0)/(a+s-0)=b/(a+s)` `M_(BD)=(b-0)/(a-s)=b/(a-s)` `M_(AC)*M_(BD) ` `b/(a+s)*b/(a-s)` `b^2/(a^2-s^2)` `b^2/-b^2` -1 multipliplication of there slope is-1 so they are perpenticular |
|
| 105. |
Find the centroids of the triangles whose vertices have the coordinates `(-7,6) (2,-2) ` and (8,5). |
|
Answer» Let A (-7,6) , B (2,-2) and C (8,5). Let the coordinates of the points A , B and C be ` (x_(1) , y_(1)), (x_(2), y_(2))` and `(x_(3), y_(3))` respectively ` therefore x_(1) = - 7 , y_(1) - 6 , x_(2) = 2 , y_(2) = - 2, x_(3) - 8 " and " y_(3) = 5` Let G (x,y) be the centroid. By dentroid formula `x = (x_(1) + x_(2) + x_(3))/(3)` ` therefore x = (-7 + 2 + 8)/(3)` ` therefore x = (3)/(3)` ` therefore x = 1` ` y= (y_(1) + y_(2) + y_(3))/(3)` ` therefore y = (6 + (-2) + 5)/(3)` `therefore y = (9)/(3)` ` therefore y = 3 ` The coordinates of the centroid are (1,3). |
|
| 106. |
ABCD is a trapezium such that AB and CD are paralleland `B C_|_C D`. If `/_A D B""=theta,""B C""=""p""a n d""C D""=""q`, then AB isequal to(1) `(p^2+q^2costheta)/(pcostheta+qsintheta)`(2) `(p^2+q^2)/(p^2costheta+q^2sintheta)`(3) `((p^2+q^2)sintheta)/((pcostheta+qsintheta)^2)`(4) `((p^2+q^2)sintheta)/(pcostheta+qsintheta)` |
|
Answer» `BD= sqrt(p^2 + q^2)` using pythagoras theorem in `/_ BCD` in `/_ABD` `(AB)/(sin theta) = (BD)/(sin(pi-(theta + alpha))` `(AB)/(sin theta) = (BD)/(sin ( theta + alpha))` `AB= (sqrt(p^2 + q^2)* sin theta)/(sin theta cos alpha + cos alpha sin theta)` `cos alpha = q/(sqrt(p^2 + q^2))` so, `AB= (sqrt(p^2 + q^2) sin theta)/(sin theta q/sqrt(p^2 + q^2)+ cos theta p/sqrt(p^2 + q^2)` `AB= (p^2 + q^2sin theta)/(q sin theta + p cos theta)` option 4 is correct |
|
| 107. |
If points A( 1,6) , B(5,2) and C(12,9) are three consective vertices of a parallelogram , then find the equation of the diagonal BD. |
|
Answer» Correct Answer - x + 3y - 8 = 0 |
|
| 108. |
If A = (3,-4) , B= (7,0) and C = (14 , -7) are the three consecutive vertices of a parallelogram ABCD , then find the slope of the diagonal BD . The following are the steps involved in solving the above problem . Arrange them is sequential order (A) `((x + 7)/(2) , (y + 0)/(2)) = ((3 + 14)/(2) , (-4-7)/(2))`. (B) The slope of BD = `(-11-0)/(10-7) = (-11)/(3)`. (C) `(x+7)/(2) = (17)/(2)` and `(y+0)/(2) - (-11)/(2)` `implies x = 10 , y = -11` `therefore D = (10 , -11)`. (D) Let the fourth vertex be D(x , y) . we know that the diagonals of a parallelogram bisect each other .A. AOCBB. DCABC. DACBD. CDAB |
|
Answer» Correct Answer - C DACB is the required sequential order. |
|
| 109. |
If the point A(x, 2) is equidistant from the points B(8, -2) and C(2, -2), find the value of x. Also, find the length of AB. |
| Answer» Correct Answer - x = 5, AB = 5 units | |
| 110. |
Find the distance between the points P(-4, 7) and Q(2, -5). |
|
Answer» The given points are P(-4, 7) and Q(2, -5). Then, `x_(1) = -4, y_(1) = 7 "and"x_(2) = 2, y_(2) = -5` `therefore PQ = sqrt((x_(2) -x_(1))^(2) + (y_(2) - y_(1))^(2))` `= sqrt({2-(-4)}^(2) + (-5-7)^(2)) = sqrt(6^(2) + (-12)^(2))` `=sqrt(36+144) = sqrt(180) = sqrt(36 xx 5) = 6sqrt(5)` units. |
|
| 111. |
If the centroid of `Delta ABC` having vertices A(a, b), B(b, c) and C(c, a) is the origin, then find the value of (a+b+c). |
|
Answer» Correct Answer - a+b+c =0 Centroid is `((a+b+c)/(3), (a+b+c)/(3))` So, `(a+b+c)/(3) = 0 rArr a+b+c = 0.` |
|
| 112. |
Find those points on x -axis, each of which is at a distance of 5 units from the point A(5, -3)` |
|
Answer» Let the required point on the x-axis be P(x, 0). Then, `PA = 5 rArr PA^(2) = 25` `rArr (x-5)^(2) + (0 +3)^(2) = 25` `x^(2) -10x +25 +9 =25 rArr x^(2) - 10x +9 =0` `x^(2) -x -9x +9 =0 rArr x(x-1) -9 (x-1) =0` `rArr (x-1) (x-9) = 0 rArr x - 1 =0 "or" x - 9=0` `rArr ` x = 1 or x = 9 Hence, the required points on the x-axis are B(1, 0) and C(9, 0). |
|
| 113. |
Find the relation between x and y such that the point P (x,y) is equidistant from the points `A(1,4)and B(-1,2).` |
|
Answer» We have `PA = PB rArr PA^(2) = PB^(2)` `rArr (x-1)^(2) + (y-4)^(2) =(x+1)^(2) + (y-2)^(2)` `rArr x^(2) + y^(2) - 2x-8y+17 = x^(2) + y^(2) +2x -4y +5` `rArr 4x+ 4y- 12 =0 rArr x+ y = 3 rArr y = 3-x` Hence, y = 3-x is the desired relation between x and y. |
|
| 114. |
Find arelation between `x`and `y`such thatthe point `(x , y)`isequidistant from the points `(3, 6)`and `(-3, 4)` |
|
Answer» `ex+ y = 5 , (1,2)` can be one of the pairs of the values of (x,y) ]Note : Student can write any pair of values of x and y satisfying the equation ` 3x + y = 45]` |
|
| 115. |
If P(x, y) is equidistant from the points A(7, 1) and B(3, 5), find the relation between x and y. |
|
Answer» Correct Answer - x-y = 2 `(7-x)^(2) + (1-y)^(2) = (3-x)^(2) + (5-y)^(2)` `rArr (49 + x^(2) -14x) + (1+y^(2)-2y) = (9+x^(2) -6x) +(25 + y^(2) - 10y)` `rArr x^(2) + y^(2) -14x -2y + 50 =x^(2) + y^(2) -6x -10y + 34` `rArr 8x-8y = 16 rArr x -y = 2` |
|
| 116. |
Find the distance of the point `(6,-6)` from the origin |
|
Answer» Let P(6, -6) be the given point and O(0, 0) be the origin. Then, `OP = sqrt((6-0)^(2) + (-6-0)^(2)) = sqrt(6^(2) + (-6)^(2))` ` = sqrt(36+36) = sqrt(72) = sqrt(36 xx 2) = 6sqrt(2)` units. |
|
| 117. |
Find a relation between x and y such that the point (x ,y) is equidistant from the points (7, 1) and (3, 5). |
|
Answer» PB=PC `sqrt((x-7)^2+(y-1)^2)=sqrt((x-3)^2+(y-5)^2)` squaring both sides and solving it `x^2-14x+49+y^2-2y+1=x^2-6x+9+y^2-10y+25` `8y+16=8x` `y=x-2` |
|
| 118. |
(Street Plan): A city has two main roads which cross each other at the centre of the city. These two roads are along the North–South direction and East–West direction. All the other streets of the city run parallel to these roads and are 200 m apart. There are about 5 streets in each direction. Using 1 cm = 200 m, draw a model of the city on your notebook. Represent the roads/streets by single lines. There are many cross– streets in your model. A particular cross–street is made by two streets, one running in the North – South direction and another in the East – West direction. Each cross street is referred to in the following maimer : If the 2nd street running in the North – South direction and 5th in the East – West direction meet at some crossing, then we will call this cross–street (2. 5). Using this convention, find:(i) how many cross – streets can be referred to as (4, 3).(ii) how many cross – streets can be referred to as (3, 4). |
|
Answer» a) are cross points b) only one cross point. |
|
| 119. |
The distance of P(3, 4) from the x-axis isA. 3 unitsB. 4 unitsC. 5 unitsD. 1 unit |
| Answer» Correct Answer - B | |
| 120. |
The distance between the points (0,5) and (-5,0) isA. 5B. `5sqrt(2)`C. `2sqrt(5)`D. 10 |
|
Answer» Correct Answer - b `:.` Distance between the points `(x_(1),y_(1))and (x_(2),y_(2)),` d`=sqrt((x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2))` Here, `x_(1)=0,y_(1)=5andx_(2)=0` `:.` Distance between the points (0,5) and (-5,0) `=sqrt((-5-0)^(2)+(0-5)^(2))` `=sqrt(25+25)=sqrt(50)=5sqrt(2)` |
|
| 121. |
The distance of the point P(-6,8) from the origin isA. 8B. `2sqrt(7)`C. 10D. 6 |
|
Answer» Correct Answer - c Distance between the points `(x_(1),y_(1))and (x_(2),y_(2))` `d=sqrt((x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2))` Here, `x_(1)=-6,y_(1)=8and x_(2)=0,y_(2)=0` `:.` Distance between P(-6,8)and orgin i.e., O(0,0), PO`=sqrt([0-(-6)]^(2)+(0-8)^(2))` `=sqrt((6)^(2)+(-8)^(2))` `=sqrt36+64=sqrt(100)=10` |
|
| 122. |
The distance between the points A(0,6) and B(0,-2) isA. 6B. 8C. 4D. 2 |
|
Answer» Correct Answer - b `:.` Distance between the points `(x_(1),y_(1))and (x_(2),y_(2)),` `d=sqrt((x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2)` Here, `x_(1)=0,y_(1)=6and x_(2)=0,y_(2)=-2` `:.` Distance between A(0,6) and B(0,-2), AB`=sqrt((0-0)^(2)+(-2-6)^(2))` `=sqrt(0+(-8)^(2))=sqrt(8^(2))=8` |
|
| 123. |
Find points on the x-axis, each of which is at a distance of 10 units from the point A(11, -8). |
| Answer» Correct Answer - (5, 0) and (17, 0) | |
| 124. |
Find the values of x for which the distance between the points P(x, 4) and Q(9, 10) is 10 units. |
| Answer» Correct Answer - x = 1 or x = 17 | |
| 125. |
Show that for any value of m the equation mx-y =4m-3 represents a straight line passing through a fixed point. What is the fixed point? |
|
Answer» `m_1x-y=4m_1-3-(1)` `m_2x-y=4m_2-3-(2)` `y=4m_1-3-m_1x` `m_2x-(3-4m+m_1x)=4m_2-3` `x(m_2-m_1)-3+4m_1=4m_2-3` `x(m_2-m_1)=4(m_2-m_1)` `x=4` `y=3-4m_1+m_1x` `y=3-4m_!+4m_1` `y=3` `P(4,3)`. |
|
| 126. |
The angle between the lines x = 10 and y = 10 is `"_______"`.A. `0^(@)`B. `90^(@)`C. `180^(@)`D. None of these |
|
Answer» Correct Answer - B The first line is parallel to the Y-axis . The second line is parallel to the X-axis . |
|
| 127. |
The points (2,3) (-1,5) and (x,-2) form a straight line , then x is `"_______"`. |
|
Answer» Correct Answer - x` = (19)/(2)` |
|
| 128. |
Find the coordinates of the point equidistant from three given points `A(5, 1), B(-3,-7)` and `C(7,-1).` |
|
Answer» Let the required point be P(x, y). Then, `PA = PB = PC rArr PA^(2) = PB^(2) = PC^(2)` `rArr PA^(2) = PB^(2) "and "PB^(2) = PC^(2)` `rArr {((x-5)^(2) +(y-1)^(2) = (x+3)^(2) + (y+7)^(2)), ((x+3)^(2) + (y+7)^(2) = (x-7)^(2) + (y+1)^(2)):}` `rArr {(x^(2)+y^(2)-10x-2y+26 =x^(2) +y^(2) +6x +14y +58), (x^(2)+ y^(2) +6x +14y +58 = x^(2) +y^(2)-14x +2y +50):}` Now, `x^(2) + y^(2) -10x -2y +26 =x^(2) + y^(2) +6x +14y +58` `rArr 16x+ 16y =-32 rArr x+y = -2 " "...(i)` And, `x^(2) +y^(2) +6x +14y +58 =x^(2) +y^(2) -14x +2y+50` `rArr 20x+12y = -8 rArr 5x +3y =-2 " "..(ii)` Multiplying (i) by 5 and subtracting (ii) from the result, we get 2y =-8 `rArr` y = -4 Putting y = -4 (i), we get x = 2. `therefore` x = 2 and y = -4. Hence, the required point is P(2, -4). |
|
| 129. |
Find the point on Y-axis which is equidistant from the points `(-5,2)and (9,-2).` |
|
Answer» Let the given points be A(-5, 2) and B(9, -2) and let the required point be P(0, y).Then, `PA = PB rArr PA^(2) = PB^(2)` `rArr (0+5)^(2) + (y-2)^(2) = (0-9)^(2) + (y+2)^(2)` `rArr 5^(2) + (y-2)^(2) = (-9)^(2) + (y +2)^(2)` `rArr 25 + (y-2)^(2) = 81 +(y +2)^(2)` `rArr (y+2)^(2) -(y-2)^(2) = 25-81` `rArr ` 8y =-56 `rArr` y = -7 Hence, the required point is P(0, -7). |
|
| 130. |
Distance of point (-3,4) from the origin is ……….A. 7B. 1C. 5D. 4 |
| Answer» Correct Answer - C | |
| 131. |
Find those points on the y-axis, each of which is at a distance of 13 units from the point A(-5, 7). |
|
Answer» Let the required point on the y-axis be P(0, y). Then, `PA = 13 rArr PA^(2) = 169` `rArr (0+5)^(2) + (y-7)^(2) = 169` `rArr y^(2)-14y +74 = 169 rArr y^(2) -14y -95 =0` `rArr y^(2) - 19y +5y -95=0 rArr y(y-19) +5(y-19) = 0` `rArr (y-19)(y+5) = 0 rArr y-19 = 0" or "y +5 = 0` `rArr y = 19 " or "y = -5` Hence, the required points on the y-axis are B(0, 19) and C(0, -5). |
|
| 132. |
Find the distance of each of the following points from the origin: (i) A (5, -12) (ii) B(-5, 5) (iii) C(-4, -6). |
| Answer» Correct Answer - (i) 13 units (ii) `5sqrt(2)` units (iii) `5sqrt(2)` units | |
| 133. |
Find thevalues of `y`for whichthe distance between the points `P(2, -3)`and `Q(10 , y)`is 10units. |
| Answer» Correct Answer - y = 3 or y = -9 | |
| 134. |
Find all possible values of x for which for distance between the points A(x, -1) and B(5, 3) is 5 units. |
| Answer» Correct Answer - x= 8 or x = 2 | |
| 135. |
Find the value of y for which the distance between the points A(3, -1)and B(11, y) is 10 units. |
|
Answer» We have AB= 10 `rArr AB^(2) = 100` `rArr (11-3)^(2) + (y+1)^(2) = 100` `rArr 8^(2) + (y+1)^(2) = 100` `rArr (y+1)^(2) = 100-64 = 36 = 6^(2)` `rArr y+1 = +-6` `rArr y+1 = 6 "or" y+1 = -6` `rArr y = 5 "or" y = -7` Hence, the required values of y and 5 and -7 |
|
| 136. |
If thepoint `P(k-1, 2)`isequidistant from the points `A(3, k)`and `B(k , 5)`, find thevalues of `k`. |
|
Answer» It is being given that P(k-1, 2) is equidistant from the points A(3, k) and B(k, 5). So, we have `PA = PB rArr PA^(2) = PB^(2)` `rArr (k-1-3)^(2) + (2-k)^(2) = (k-1-k)^(2) + (2-5)^(2)` `rArr (k-4)^(2) + (2-k)^(2) = (-1)^(2) + (-3)^(2)` `rArr 2k^(2)-12k +20 = 1+9 rArr 2k^(2) -12k +10=0` `rArr k^(2)-6k+5 = 0rArr k^(2) - 5k-k+5=0` `rArr k(k-5) - (k-5) = 0 rArr (k-5) (k-1) = 0` `rArr k= 1 "or" k = 5` Hence, the required values of k are 1 and 5 |
|
| 137. |
If the point C(-2, 3) is equidistant from the points A(3, -1) and B(x, 8), find the values of x. Also, find the distance BC. |
|
Answer» Correct Answer - x=2 or x = -6, BC =`sqrt(41)` units `AC = BC rArr AC^(2) = BC^(2)` `rArr (3+2)^(2) + (-4)^(2) = (x+2)^(2) + 5^(2)` `rArr (x+2)^(2) = 16 rArr x+ 2 = 4"or" x + 2 = -4` `rArr x= 2 "or" x= -6.` Distance between `B(2, 8) "and"C(-2, 3) = sqrt(4^(2) +5^(2)) = sqrt(41)` units. Distance between `B(-6, 8) " and " C(-2, 3) = sqrt(4^(2) + (-5)^(2)) = sqrt(41)` units. |
|
| 138. |
If the point P(2, 2) is equidistant from the points A(-2, k) and B(-2k, -3), find k. Also, find the length of AP. |
|
Answer» Correct Answer - `(k =-3 rArr AP = sqrt(41)), (k = -1 rArr AP = 5)` `PA = PB rArr PA^(2) = PB^(2)` `rArr (2+2)^(2) + (2-k)^(2) = (2+2k)^(2) + (2+3)^(2)` `rArr (2-k)^(2) - (2 +2k)^(2) = 25-16 =9` `rArr (4+k^(2)-4k) -(4+4k^(2) +8k) = 9` `rArr 3k^(2) + 12k + 9=0` `rArr k^(2) + 4k +3 =0 rArr (k +3)(k+1) = 0` `rArr k = -3 "or" k =-1` Case 1. When k = -3, we have AP = distance between A(-2, -3) and P(2, 2) = `sqrt(41)` units. Case 2. When k =-1, we have AP = distance between A(-2, -1) and P(2, 2) = sqrt(25) = 5 units. |
|
| 139. |
A thief runs with a uniform speed of 100 m/min. After one minute a policeman runs after the thief to catch him. He goes with a speed of 100 m/min in first minute and increases his speed by 10 m/min every succeeding minute. After how many minutes the policeman will catch the thief. |
|
Answer» distance travelled by thief = `100m/min xx 1 min` `= 100m` distance between thief andd police men `= 100m` after 2 mins 3 mins ; `d = 100-10 = 90metre` `120-> 100 , 4 min` `d = 70-20 = 70 m` `130-> 100 ; 5min` `d = 70-30 = 40 m` `140 -> 100; 6min` `d-> 40-40 = 0m` so, `40m` after 6 mins policeman will catch the thief `= 10+20+30+40----= 100metre` 6 min answer |
|
| 140. |
Find the ratio in which the point `P(x,2)` divides the line segment joining the points `A(12,5) and B(4, -3)`. Also find the value of `x`. |
|
Answer» Let the required ration be k:1. Then, by section formula, the coordinates of P are `P((4k+12)/(k+1), (-3k +5)/(k+1))` But, this point is given as P(x, 2). `therefore (-3k +5)/(k+1) = 2 rArr -3k + 5 = 2k +2 rArr 5k = 3 rArr k = (3)/(5).` So, the required ratio is 3:5. Putting `k = (3)/(5)` in P, we get `x = ({4 xx (3)/(5) + 12})/(((3)/(5) +1)) = (72)/(8) = 9.` Hence, x = 9. |
|
| 141. |
Find the distance between A(2,3) and B(4,1). |
|
Answer» Soppose the coordinates of point A are `(X_(1),y_(1))` and of point B are ` (x_(1) , y_(2))` ` x_(1) = 2 y_(1) = 3 , x_(2) = 4 and y_(2) = 1` By distance formula , `AB = sqrt((x_(2) - x_(1))^(2) + (y_(1) - y_(2))^(2))` `= sqrt((4 - 2)^(2) + (1 - 3)^(2))` ` = sqrt(2^(2) + (-2)^(2))` ` = sqrt( 4 + 4 )` ` = sqrt(8) = 2 sqrt(2)` Distance between point A and B is ` 2sqrt(2)` . |
|
| 142. |
Determine whether the points A(1, -3) , B(2,-5) and C(-4,7) are collinear or not . |
|
Answer» A(1, -3) , B(2,-5) and C(-4,7) By distance formula, `d(A),(B) = sqrt((2-1)^(2)+[-5-(-3)]^(2))` `=sqrt((1)^(2) + (-5+3)^(2))` `= sqrt((1)^(2)+(-5+3))^(2))` ` =sqrt( 1+(-2)^(2))` ` = sqrt(1+4)` `= sqrt(5) `…(i) `d(B,C=)= sqrt((-4) -2)^(2) + [7-(-5)]^(2))` `= sqrt((-6)^(2) + (12)^(2))` ` = sqrt(38 + 144)` ` = sqrt(180)` ` sqrt(2xx2xx3xx3xx5)` ` = 6 sqrt(5)` ...(2) `d(A,C) = sqrt((-4-1)^(2) + [7-(-3)]^(2))` ` = sqrt((-5)^(2) + (10)^(2))` `sqrt( 25 + 100) = sqrt(125)` `sqrt(5xx5xx5xx5)` ` = 5sqrt(5)` ...(3) Adding (1) and (3) `d (A , B) + d(A,C)= sqrt(5) + 5sqrt(5) = 6 sqrt(5)` ` therefore A(A,B + d(A,C) = d (B,C)" "` ...From (2)] ...(4) `therefore ` Point A,B and C are collinear. Point A(1,-3) , (2,-5) and C(-4,7) are colliear. |
|
| 143. |
Show that the line joining the points A(4,8) and B(5,5) is parallel to the line joining the points C(2,4) and D(1,7). |
|
Answer» A(4,8),B(5,5),C(2,4) and C(1,7) Slope of line ` AB = (5-8)/(5-4)` `= (-3)/(1) = - 3` Slope of line `CD = (7-4)/(1-2) ` `= (3)/(-1) = -3` ` therefore ` slope of line AB = slope of line CD Parallel lines have equal slopes . ` therefore " line AB "||` line CD . |
|
| 144. |
Using slope concept , determine whether R(1,-4), S(-2,2) and T(-3,4) are collinear or not . |
|
Answer» R(1,-4),S(-2,2) and T(-3,4) Slope of line `RS = (2-(-4))/(-2-1)` `= (2+4)/(-3)` ` = (6)/(-3) = - 2` Slope of line `ST = (4-2)/(-3-(-2))` ` = (2)/(-1) = -2` Slope of line RS = slope of line ST and S is the common point . ` therefore ` points R,S and T are collinear. Points R (1,-4), S (-2,2),and T (-3,4) are collinear. |
|
| 145. |
Show that the points (2,0), (-2,0) and (0,2) are the vertices of a triangle. Also state with reason the type of the triangle . |
|
Answer» Let A(2,0), B(-2,0) and C(0,2) By distance formula `AB = sqrt((x_(2) - x_(1))^(2) + (Y_(2) - y_(2))^(2))` ` = sqrt((-2-2)^(2) + (0- 0)^(2))` ` = sqrt((-4)^(2) + (0)^(2))` ` sqrt(16)` ` = 4 ` Similary , ` BC = 2sqrt(2)" and " AC = 2sqrt(2)` For any triangle Sum of two sides will always be greter then the third side ` therefore 2sqrt(2) + 2sqrt(2) gt 4 ` i.e. ` 4sqrt(2) gt 4 ` and also ` 4 + 2 sqrt(2) gt 2sqrt(2)` ` therefore A ,B C ` are the vertices of a triangle Now, ` BC^(2) + AC^(2) = (2sqrt(2)^(2)+ 2sqrt(2))^(2)` ` therefore BC^(2) + AC^(2) = 16` ...(1) `AB^(2) = 4^(2)` ` therefore AB^(2) = 16` ...(2) ` therefore BC^(2) + AC^(2) = AB^(2)" "` [from (1) and (2)] ` therefore ` by converse, of Pythagoras theorem ` Delta ABC ` is a right angled triangle and also BC = AC ` therefore Delta ABC ` is an isosceles right angled triangle. |
|
| 146. |
Wrtie down the coordinates of centroid of the triangle whose vertices (4,7),(8,4) and (7,11). |
|
Answer» The coordinates of centroid of the triangle whose vertices (4,7),(8,4) and (7,11) are `((19)/(3),(22)/(3))` |
|
| 147. |
Find the point on the X-axis which is equidistant from (-3,4) and B(1,-4). |
|
Answer» A(-3,4) and B(1,-4) Let P be the point on X-axis equidistant from points A and B . ` therefore PA = PB ` ….(1) P lies on X-axis ` therefore ` its y coordinate is 0. Let P(x,0) Now , PA = PB …[From (1)] By distance formula , ` ([x-(-3)]^(3) + (0-4)^(2))=sqrt((x-1)^(2) + [0-(-4)]^(2))` ` therefore sqrt((x +3)^(2) + (-4)^(2))= sqrt((x-1)^(2) + (4)^(2))` Squaring both the sides , `(x + 3)^(2) + 16 = (x-1)^(2) + 16` ` therefore x^(2) + 6x + 9 = x^(2) - 2x + 1` ` therefore 6x + 2x= 1 - 9 ` ` therefore 8x = - 8` ` therefore x = - (8)/(8) therefore x = - 1` ` therefore P(-1,0) ` The coordinates of point on X-axis equidistant from points A and B are (-1,0) . |
|
| 148. |
A(h,-6),B(2,3) and C(-6,k) are the coordinates of vertices of a triangle whose centroid is G(1,5). Find h and k. |
|
Answer» A(h,-6) , B (2,3),C(-6,k) and G(1,5) . Suppose A has coordintes ` (X_(1),y_(1)) ` , B has coodintes ` (X_(2),y_(2))` C has coordinates `(X_(3) ,y_(3))` and Gn has coordinates (x,y). Here , `x_(1) = h,y_(1) = - 6, x_(2) = 2 , y_(2) = 3 , x_(3) = - 6,y_(3) = k , x = 1` and y=5. By centroid formula , ` x= (x_(1) + x_(2) + x_(3))/(3)` `therefore 1 = (h + 2+ (-6))/(3)` ` therefore 3 = h-4` ` therefore = 3+ 4 ` ` therefore h = 7` ` y=(y_(1) + y_(2) + y_(3))/(3) ` ` therefore 5 = (-6 + 3 + k)/(3)` ` therefore 15 = - 3 + k` ` therefore k = 15+ 3 ` ` therefore k = 18` ` h = 7 ad kk = 18` . |
|
| 149. |
Show that the points A(1,2),B(1,6), `C(1 + 2sqrt(3),4) ` are the vertices of an equilateral triangle . |
|
Answer» `A(1,2),B(1,6),C(1 + 2sqrt(3),4)` By distance formula , ` AB = sqrt((1 - 1)^(2) + (6-2)^(2))` ` therefore AB = sqrt(0^(2) + 4^(2))` ` therefore AB= sqrt(4^(2))` ` therefore AB = 4 ` `BC = sqrt((1 + 2sqrt(3)-1)^(2) + (4 -6)^(2))` ` therefore BC= sqrt((2sqrt(3)^(2) + )(-2)^(2))` ` therefore BC = sqrt(12 + 4)` ` therefore BC = sqrt(16) ` `therefore BC = 4` `therefore AC = sqrt((1 + 2sqrt(3)-1)^(2) + (4-2)^(2))` ` therefore AC = sqrt((2sqrt(3))^(2) + (2)^(2))` ` therefore AC = sqrt(12 + 4)` ` therefore AC = sqrt(16)` ` therefore AC = 4 ` ` therefore AB = BC = AC ` ` therefore Delta ABC ` is an equilateral triangle .i.e. the points A , B and C ar the vertices of an equilateral triangle . |
|
| 150. |
If `P(-5, -3), Q(-4, -6), R(2, -3)`and `S(1, 2)`are thevertices of a quadrilateral `P Q R S`, find itsarea. |
| Answer» Correct Answer - 28 sq units | |