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(3) √26

Distance = \(\sqrt{(5-0)^2+(-1-0)^2}\)

= \(\sqrt{5^2+1^2}=\sqrt{26}\)

The sum of two numbers x and y is 75.

∴ x + y = 75

Given equation = 5x + 3y – 7 = 0 

x = 2 – α and y = 2 + α 

∴ 5(2 – α) + 3(2 + α) – 7 = 0 

⇒ 10 – 5α+ 6 + 3α – 7 = 0 

⇒ 9 – 2α = 0 

⇒ 2α = 9 ⇒ α = 9/2 = 4.5 

Given equation 3x – 2y + 6 = 0 

x = 2β + 1 and y = β – 1 

∴ 3(2β + 1)- 2(β – 1) + 6 = 0 

⇒ 6β + 3 – 2β + 2 + 6 = 0

⇒ 4β + 11 = 0 ⇒ β = -11/4

∴ α + β = 9/2 - 11/4 = \(\frac {18-11}{4} = \frac {7}{4}\)

Correct option is (A) \(\frac{-c}{a}\)

Given line is ax+by+c = 0

\(\Rightarrow ax+by=-c\)

\(\Rightarrow\frac{ax}{-c}+\frac{by}{-c}=1\)

\(\Rightarrow\frac{x}{\frac{-c}{a}}+\frac{y}{\frac{-c}{b}}=1\)

By comparing with intercept form of line, we get

X-intercept \(=\frac{-c}a\) and Y-intercept \(=\frac{-c}b\)

Hence, X-intercept of the line ax+by+c = 0 is \(\frac{-c}{a}.\)

Correct option is A) \(-\cfrac{c}a\)

(b) (2, 2)

Let the point be P whose abscissa = ordinate = a. 

∴ P ≡ (a, a) 

Given, PA = PB 

⇒ (a + 1)² + a² = a² + (a – 5)² 

⇒ 2a² + 2a + 1 = 2a² – 10a + 25 

⇒ 12a = 24 ⇒ a = 2. 

∴ The point is (2, 2).

The points (ii) (0,13) ; (v) (0,- 8) lie on Y – axis. 

The points (iv) (- 2, 0), (vi) (7, 0) lie on X – axis. 

The point (vii) (0, 0) lie on both X – axis and Y – axis. 

The points (i) (- 5, – 8); (iii) (4, – 2) do not lie on any axis.

Given points (3, 4) and (4, 3) having x and y coordinates are equal.

That line is Y – axis.

The general form of points lying on X – axis is (x, 0).

All the above points lie on Y – axis, since the X – coordinate of all these points is zero.

Correct option is (C) 6

Slope of line through points (2, -7) and (x, 5) is

\(m=\frac{y_2-y_1}{x_2-x_1}\)

\(=\frac{5-(-7)}{x-2}=\frac{12}{x-2}\)

But given that slope is 3.

\(\therefore\) \(\frac{12}{x-2}=3\)

\(\Rightarrow x-2=\frac{12}3=4\)

\(\Rightarrow x=4+2=6\)

Correct option is C) 6

Correct option is (A) y = 0

The equation of X–axis is y = 0

Correct option is  A) y = 0

Correct option is (C) (0, –2), (0, 2)

Points which lies on Y–axis has x-coordinate zero.

\(\because\) (0, – 2) & (0, 2) have x-coordinate zero.

Thus, both points (0, –2) & (0, 2) lie on Y-axis.

C) (0, – 2), (0, 2)

Correct option is (B) 4

Given that points (7, -2), (5, 1), (3, k) are collinear.

\(\therefore\) Area of formed triangle is zero.

\(\therefore\frac12|7(1-k)+5(k-(-2))+3(-2-1)|=0\)

\(\Rightarrow|7-7k+5k+10-9|=0\)

\(\Rightarrow|-2k+8|=0\)

\(\Rightarrow-2k+8=0\)

\(\Rightarrow2k=8\)

\(\Rightarrow k=\frac82=4\)

Correct option is B) 4

Correct option is (B) √2

Distance between points \(A(cos\,\theta,sin\,\theta)\;and\;B(-sin\,\theta,cos\,\theta)\) is \(AB=\sqrt{(-sin\,\theta-cos\,\theta)^2+(cos\,\theta-sin\,\theta)^2}\)

\(=\sqrt{sin^2\,\theta+cos^2\,\theta+2\;sin\,\theta\,cos\,\theta+cos^2\,\theta\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+sin^2\,\theta-2\;sin\,\theta\,cos\,\theta}\)

\(=\sqrt{2(sin^2\,\theta+cos^2\,\theta)}\)

\(=\sqrt{2\times1}\)           \((\because sin^2\,\theta+cos^2\,\theta=1)\)

\(=\sqrt{2}\) units

Correct option is B) √2

Correct option is (A) \((\frac{2}{3},2)\)

Centroid of (-1, 3), (6, -3) and (-3, 6) is

\(\left(\frac{-1+6+(-3)}3,\frac{3+(-3)+6}3\right)\)

\(=(\frac{6-4}3,\frac{6}3)=(\frac{2}{3},2)\)

Correct option is A) \((\cfrac{2}{3},2)\)

Correct option is (B) x–coordinate

The distance of a point from Y-axis is called its x-coordinate or abscissa.

Correct option is B) x – coordinate

Correct option is (B) 2 + √2

Given vertices of triangle are \(A(0,0),B(1,0)\;and\;C(0,1).\)

\(\therefore\) \(AB=\sqrt{(1-0)^2+(0-0)^2}\)

\(=\sqrt{1+0}\)

= 1 units,

\(BC=\sqrt{(0-1)^2+(1-0)^2}\)

\(=\sqrt{1+1}\)

= \(\sqrt2\) units

and \(AC=\sqrt{(0-0)^2+(1-0)^2}\)

\(=\sqrt{0+1}\)

= 1 units

\(\therefore\) Perimeter of formed triangle ABC \(=AB+BC+AC\)

\(=(1+\sqrt2+1)\) units

\(=2+\sqrt2\) units

Correct option is B) 2 + √2

Correct option is (A) Q₁

\(\because\) x < 0 and y > 0

\(\Rightarrow\) -x > 0 and y > 0

\(\Rightarrow\) (-x, y) lies in \(Q_1\) quadrant.

Correct option is  A) Q1 

Correct option is (C) 6

A triangle has 3 vertices and 3 sides.

\(\therefore\) A triangle has 6 elements.

Correct option is C) 6

Correct option is (A) 6

We have A = (4, 2), B = (1, y) and AB = 5

\(\Rightarrow\sqrt{(1-4)^2+(y-2)^2}=5\)

\(\Rightarrow(-3)^2+(y-2)^2=5^2=25\)        (By squaring both sides)

\(\Rightarrow(y-2)^2=25-9=16^2=4^2\)

\(\Rightarrow y-2=4\;or\;y-2=-4\)

\(\Rightarrow y=4+2=6\;or\;y=-4+2=-2\)

Correct option is A) 6

Correct option is (B) perpendicular

In the coordinate plane the axes are perpendicular to each other.

Correct option is B) perpendicular

Correct option is (D) sec θ

Distance of point \(A(tan\,\theta,1)\) from origin O(0, 0) is

\(AO=\sqrt{(0-tan\,\theta)^2+(0-1)^2}\)   (By distance formula)

\(=\sqrt{tan^2\,\theta+1}\)

\(=\sqrt{sec^2\,\theta}\)                 \((\because1+tan^2\,\theta=sec^2\,\theta)\)

\(=sec\,\theta\)

Hence, distance of the point \((tan\,\theta,1)\) from origin is \(sec\,\theta.\)

Correct option is D) secθ

Correct option is (A) x – y + k = 0

Straight line which is parallel to x – y + 1 = 0 is \(x-y+k=0.\)

Correct option is A) x – y + k = 0

Correct option is (A) (5, 4)

x = Obscissa = Distance of point from Y-axis = 5,

y = Ordinate = Distance of point from X-axis = 4.

Thus, the point is (5, 4).

Correct option is  A) (5, 4)

Correct option is (D) 20

Given that \(\overline{AB}\) = 12 units where A = (8, 3) and B = (x, 3)

\(\therefore\sqrt{(x-8)^2+(3-3)^2}=12\)        (By distance formula)

\(\Rightarrow(x-8)^2=12^2\)

\(\Rightarrow x-8=12\;or\;x-8=-12\)

\(\Rightarrow x=12+8=20\;or\;x=8-12=-4\)

Correct option is D) 20

The point where coordinate axes intersect is known as origin O(0, 0)

Correct option is (C) A and B

The equation of Y-axis is x = 0.

The equation of line parallel to Y-axis is x = c, where c is constant.

\(\therefore\) x = 2017 is a line parallel to Y-axis.

Also, slope of Y-axis is not defined.

\(\therefore\) Slope of line parallel to Y-axis is not defined.

\(\therefore\) Slope of line x = 2017 is not defined.

Hence, option (A) and (B) both are correct.

Correct option is C) A and B

Correct option is (D) (3, 8)

y-coordinate of a point = Distance of that point from X-axis

\(\Rightarrow y=8\)                         (Given)

x-coordinate of a point = Distance of that point from Y-axis

\(\Rightarrow x=3\)

\(\therefore\) Point whose distance from X-axis is 8 units and distance from Y-axis is 3 units is denoted by (x, y) = (3, 8).

Correct option is D) (3, 8)

Given: Origin and a point (7, 4). 

Distance of a point (x, y) from the origin is 

\(\sqrt{x^2+y^2}\)

= \(\sqrt{7^2+4^2}\) 

= \(\sqrt{49+16}\) 

= \(\sqrt{65}\) units.

Correct option is (D) 4

Given that distance of point A(3, p) from origin O(0, 0) is 5 units.

\(\therefore AO=5\)

\(\Rightarrow AO^2=5^2=25\)

\(\Rightarrow(0-3)^2+(0-p)^2=25\)    \((\because AO=\sqrt{(0-3)^2+(0-p)^2})\)

\(\Rightarrow9+p^2=25\)

\(\Rightarrow p^2=25-9=16\)

\(\Rightarrow p=\pm4\)

\(\Rightarrow p=4\;or\;p=-4\)

Correct option is D) 4

Correct option is (D) (-7, 0)

On X-axis, y-coordinate of any point is zero.

\(\therefore\) Put y = 0 in equation y = x+7 to get the intersecting point with X-axis.

\(x+7=0\)

\(\Rightarrow x=-7\)

Hence, straight line y = x+7 intersect X-axis at point (-7, 0).

Correct option is D) (-7, 0)

Correct option is (C) 1

Given line is 3x – by + 7 = 0

\(\Rightarrow by=3x+7\)

\(\Rightarrow y=\frac3bx+\frac7b\)

\(\therefore\) Slope of given line is \(\frac3b\)

\(\therefore\) Slope of line parallel to given line is \(\frac3b.\)

But given that the slope of line parallel to \(3x-by+7=0\) is 3.

\(\therefore\) \(\frac3b\) = 3

\(\Rightarrow b=\frac33=1\)

Correct option is C) 1

Correct option is (C) - b/a

Slope of line joining the points (a, 0) and (0, b) is \(m=\frac{y_2-y_1}{x_2-x_1}\)

\(=\frac{b-0}{0-a}=\frac{b}{-a}=\frac{-b}{a}\)

Correct option is C) - b/a

Given points = (- 4, 0) and (6, 0). 

These two points lie on the X – axis.

∴ Distance between them = [x₂ – x₁ | = |6 – (-4)]

= 6 + 4 = 10 units.

Correct option is (C) X-axis

Any point on X-axis has y-coordinate zero.

Since, y-coordinate of all given points (-4, 0), (2, 0), (6, 0) & (-8, 0) is zero.

\(\therefore\) All given points (-4, 0), (2, 0), (6, 0) & (-8, 0) lie on X-axis in the coordinate axes.

Correct option is C) X-axis

Correct option is (C) \(\frac{1}{2}\)

Given line is 2x + y + 4 = 0

\(\Rightarrow y=-2x-4\)

By comparing with y = mx+c, we obtain

\(m=-2\)

Hence, slope of the line 2x + y + 4 = 0 is m = -2.

\(\therefore\) Slope of line perpendicular to line 2x+y+4 = 0

\(=\frac{-1}{\text{slope of line 2x+y+4}}\)

\(=\frac{-1}{-2}=\frac12\)

Hence, the slope of the line perpendicular to the line 2x+y+4 = 0 is \(\frac{1}{2}.\)

Correct option is D)\(\cfrac{1}2\)

Given points are (- 4, 0), (2, 0), (6, 0), (- 8, 0). 

All these points have their y-coordinates = 0

∴ These points lie on X-axis.

Slope = \(\frac{y_2-y_1}{x_2-x_1}\)

\(= \frac{-2 + 2}{-6-3} \)

= 0

Slope = \(\frac{y_2-y_1}{x_2-x_1}\)

\(=\frac{0-4}{4-0}\)

\(=\frac{-4}{4}\)

= -1

Slope = \(\frac{y_2-y_1}{x_2-x_1}\)

= \(\frac{b-0}{0-a}\)

= \(\frac{-b}{a}\)

Correct option is (C) 0

Given that line 4x + 5y = k passing through the origin (0, 0).

By putting x = 0 & y = 0 in 4x + 5y = k, we get

\(k=4\times0+5\times0\)

\(=0+0=0\)

Hence, if the line 4x+5y = k passing through the origin, then k = 0.

Correct option is C) 0

Correct option is (A) 5

The slope of line joining points (2, 1) & (3, 6) is

\(m=\frac{y_2-y_1}{x_2-x_1}=\frac{6-1}{3-2}\)

\(=\frac{5}{1}=5\)

Hence, slope of the line joining points (2, 1) & (3, 6) is 5.

Correct option is A) 5

Correct option is (C) y = x

Let the line be y = mx+c   ________________(1)

Since, line is passing through the point (0, 0)

\(\therefore\) By putting x = 0 & y = 0 in equation of line (1), we obtain

\(0=m\times0+c\)

\(\therefore c=0-0=0\)

Put c = 0 in equation of line (1), we get

\(y=mx\)             ________________(2)

Equation (2) represent the family of lines which passes through origin (0, 0).

Since, lines in options (A), (B) & (D) have constant terms.

Thus these lines does not pass through origin (0, 0).

Put m = 1 in equation (2), we get

\(y=x\)

Hence, \(y=x\) is a line passing through the point (0, 0).

Correct option is C) y = x

Correct option is (A) √5

Distance between the points A(3, 2) and B(2, 4) is

\(AB=\sqrt{(2-3)^2+(4-2)^2}\)

\(=\sqrt{(-1)^2+2^2}\)

\(=\sqrt{1+4}\)

\(=\sqrt{5}\) units

Correct option is A) √5

Correct option is (A) 5/6

Given line is 5x – 6y + 10 = 0

\(\Rightarrow6y=5x+10\)

\(\Rightarrow y=\frac56x+\frac{10}6\)

\(\Rightarrow y=\frac56x+\frac53\)

By comparing with y = mx+c, we get

\(m=\frac{5}{6}\)

Hence, slope of given line is \(\frac{5}{6}.\)

Correct option is A) 5/6

The points (3,a) lies on the line 2x - 3y = 5.

If point (3,a) lies on the line 2x - 3y = 5, then 2x - 3y = 5

\(\Rightarrow\) (2 x 3) - (3 x a) = 5

\(\Rightarrow\) 6 - 3a = 5

\(\Rightarrow\) 3a = 1

\(\Rightarrow\) a = \(\frac{1}{3}\)

Hence, the value of a is \(\frac{1}{3}\).

Co-ordinate Geometry is that branch of geometry in which two numbers, called co-ordinates are used to indicate the position of a point in a plane and which make use of algebraic methods in the study of geometric figures.

D) Rene Descartes

If (x₁, y₁), (x₂, y₂), (x₃, y₃) (x₁, y₁) (x₂, y₂), (x₃, y₃) are the vertices of a triangle then its area is given by

Area = [1/2 (x₁(y₂ −y₃)+x₂ (y₃ −y₁ )+x₃ (_y1 −y₂ ))]

Area = \(\frac{1}2\)[(0(5-4)+0(4-1)+3(1-5)]

⇒ \([\frac{1}{2}(-12)]\)

⇒ 6 sq. units