Coordinate Geometry — Part 2 MCQs with Explanations

$$\eqalign{ & \Rightarrow {\text{mid point co - ordinate}}\left( {\frac{{ - 5 + x}}{2},\,\frac{{y + 3}}{2}} \right) \cr & \Rightarrow \frac{{ - 5 + x}}{2} = - 2 \cr & \Rightarrow - 5 + x = - 4 \cr & \Rightarrow x = 1 \cr} $$

$$\eqalign{ & \frac{{k:1}}{{{\text{A}}\left( { - 2,\,1} \right)\,\,\,\,\,\,\,\,\,\,{\text{C}}\left( {x,\,y} \right)\,\,\,\,\,\,\,\,\,\,{\text{B}}\left( {2,\,3} \right)}} \cr & \Rightarrow x = \frac{{ - 2k + 2}}{{k + 1}} \cr & {\text{At }}y{\text{ - axis}}\,\boxed{x = 0} \cr & \therefore 0 = \frac{{ - 2k + 2}}{{k + 1}} \cr & \Rightarrow \boxed{k = 1} \cr & {\text{Ratio}} = 1:1 \cr} $$

3x + 4y = 10 . . . . . . . (i) -x + 2y = 0 . . . . . . . (ii) On solving both the equation 3(2y) + 4y = 10 10y = 10 y = 1 ∴ x = 2 × 1 = 2 ∴ Solution (a, b) = (2, 1) ∴ a + b = 2 + 1 = 3

Slope (m 1 ) for the line PQ $$ = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{0 - 3}}{{4 + 1}} = \frac{{ - 3}}{5}$$ Now, required equation of the line passing through (-1, 3) and having x-intercept of 4 $$\eqalign{ & y = m\left( {x - a} \right) \cr & \Rightarrow y = - \frac{3}{5}\left( {x - 4} \right) \cr & \Rightarrow 5y = - 3x + 12 \cr & \Rightarrow 3x + 5y = 12 \cr} $$

Parallelogram ABCD AD || BC Slope of line AD (m) = Slope of line BC Slope of line BC $$ = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{4 - 3}}{{ - 1 + 2}} = 1$$ Slope of AD = 1 Equation of line AD ⇒ y - y 1 = m(x - x 1 ) x 1 = 5, y 1 = 0 y - 0 = 1(x - 5) y = x - 5

$$\eqalign{ & ax - 4y = - 6 \cr & \Rightarrow 4y = ax + 6 \cr} $$ $$\eqalign{ & y = \frac{a}{4}x + \frac{6}{4} \cr & \frac{a}{4} = \frac{{ - 3}}{2} \cr & \boxed{a = - 6} \cr} $$

2x + 5y - 12 = 0 x + y = 3 at y = 0 (3, 0), (6, 0) at x = 0 (0, 2.4)(0, 3) Area of Δ = $$\frac{1}{2}$$ × 3 × 2 = 3

Slope of the line passing through the point (-5, 1) & (-2, 0) $${m_1} = \frac{{0 - 1}}{{ - 2 - \left( { - 5} \right)}} = \frac{{ - 1}}{3}\,\,\,\,\,\left[ {m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right]$$ Slope of the 1 st line $$ = \frac{{ - 1}}{{{m_1}}} = \frac{{ - 1}}{{\left( {\frac{{ - 1}}{3}} \right)}} = 3$$ (∴ If 2 lines are ⊥ then their slope m 1 × m 2 = -1)

18x - 72y + 13 = 0 7x - my - 17 = 0 There is no solution, means they are parallel $$\eqalign{ & \frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \cr & \Rightarrow \frac{{18}}{7} = \frac{{72}}{m} \cr & \Rightarrow m = 7 \times 4 \cr & \Rightarrow m = 28 \cr} $$

Slope = tanθ Here, the angle made by the given line is more than 90° ∴ θ > 90° but θ < 180° Since, tanθ in second quadrant is negative. ∴ Slope of the given line is negative.

Area bounded by lines x = 0, y = 0, x + y = 1, 2x + 3y = 6 is $$\eqalign{ & \Rightarrow 2x + 3y = 6 \cr & \Rightarrow \frac{{2x}}{6} + \frac{{3y}}{6} = 1 \cr & \Rightarrow \frac{x}{3} + \frac{y}{2} = 1 \cr} $$ $$\eqalign{ & {\text{On drawing the lines on graph,}} \cr & \therefore {\text{Area bounded}} \cr & = \frac{1}{2} \times 3 \times 2 - \frac{1}{2} \times 1 \times 1 \cr & = 3 - \frac{1}{2} \cr & = \frac{5}{2} \cr & = 2\frac{1}{2}{\text{ sq}}{\text{. units}} \cr} $$

\[\begin{array}{*{20}{c}} {{\text{Point }}\left( { - 3,\,0} \right){\text{ and }}\left( {0,\,6} \right)} \\ {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x_1}{y_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x_2}{y_2}} \end{array}\] $$\eqalign{ & {\text{Equation of line}} \cr & y - {y_1} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right) \cr & y - 0 = \frac{{6 - 0}}{{0 - \left( { - 3} \right)}}\left( {x - \left( { - 3} \right)} \right) \cr & y = \frac{6}{3}\left( {x + 3} \right) \cr & y = 2x + 6 \cr} $$

$$\eqalign{ & 3{x_{ \times 3}} - 2{y_{ \times 3}} = {8_{ \times 2}} \to \left( {\text{i}} \right) \cr & \underline {4{x_{ \times 2}} + 3{y_{ \times 2}} = {5_{ \times 2}}} \to \left( {{\text{ii}}} \right) \cr & 17x = 34 \cr & x = 2 \cr & y = - 1 \cr & \left( {2\alpha - \beta } \right) = \left\{ {2\left( 2 \right) - \left( { - 1} \right)} \right\} = 4 + 1 = 5 \cr} $$

$$\eqalign{ & {\text{For no - solution,}} \cr & \frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}} \cr & 2x - ky + 7 = 0{\text{ and }}6x - 12y + 15 = 0 \cr & \therefore \frac{2}{6} = \frac{{ - k}}{{12}} \cr & k = 4 \cr} $$

2 x . 2 y = 8 ⇒ 2 x + y = 2 3 On comparing both sides x + y = 3 . . . . . . . (i) Also, 9 x . 3 y = 81 ⇒ 3 2x . 3 y = 3 4 ⇒ 3 2x + y = 3 4 On comparing both sides 2x + y = 4 . . . . . . . (ii) On solving equation (i) and (ii) x = 1 y = 2 ∴ (x, y) = (1, 2)

$$\eqalign{ & 2x + 3y = 12 \cr & \Rightarrow \frac{{2x}}{{12}} + \frac{{3y}}{{12}} = 1 \cr & \Rightarrow \frac{x}{6} + \frac{y}{4} = 1 \cr} $$ ∴ Straight line 2x + 3y = 12, passes through 1 st , 2 nd and 4 th quadrant.

On plotting the points on graph, $$\therefore {\text{Area of }}\Delta = \frac{1}{2} \times 5 \times 8 = 20{\text{ sq}}{\text{. units}}$$

A(3, -2), B(1, 4), C(-2, x) Condition for collinearity of three points 0 = x 1 (y 2 - y 3 ) + x 2 (y 3 - y 1 ) + x 3 (y 1 - y 2 ) 0 = 3(4 - x) + 1(x + 2) + (-2)(-2 - 4) 0 = 12 - 3x + x + 2 + 12 2x = 26 x = 13 Alternate Solution: For collinear condition slope will be same $$\eqalign{ & \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{{y_3} - {y_2}}}{{{x_3} - {x_2}}} \cr & \frac{{4 + 2}}{{1 - 3}} = \frac{{x - 4}}{{ - 2 - 1}} \cr & \frac{6}{{ - 2}} = \frac{{x - 4}}{{ - 3}} \cr & 9 = x - 4 \cr & x = 13 \cr} $$

2x - 3y + 6 = 0 y = 0 ⇒ 2x - 3 × 0 = -6 2x = -6 x = -3 y = 0 y = 0 ⇒ 4x + 0 = 16 x = 4 ; y = 0 $$\eqalign{ & 2x - 3y = - 6\,\,\, * 2 \cr & \underline {4x + y = 16\,\,} \,\,\,\,\, * 1 \cr} $$ 4x - 6y = -12 . . . . . . (i) 4x + y = 16 . . . . . . (ii) Solve equation (i) and (ii) y = 4 ; x = 3 $$\Delta {\text{ABC}} = \frac{1}{2} \times \left( {4 + 3} \right) \times 4 = 14$$

$$\eqalign{ & \left( {x,\,y} \right) = \left( {\frac{{{m_1}{x_2} + {m_2}{x_1}}}{{{m_1} + {m_2}}},\,\frac{{{m_1}{y_2} + {m_2}{y_1}}}{{{m_1} + {m_2}}}} \right) \cr & 5 = \frac{{2 \times 0 + 5 \times x}}{{2 + 5}} \cr & 35 = 5x \cr & x = 7 \cr & - 2 = \frac{{2 \times y + 5 \times 0}}{{2 + 5}} \cr & - 14 = 2y \cr & y = - 7 \cr & x = 7,\,y = - 7 \cr} $$

Given, Equation of the lines: x - y = -1 . . . . . . (i) 3x - 2y = 0 . . . . . . (ii) From equation (i) & (ii) Intersecting co-ordinate (x 1 , y 1 ) = (2, 3) m = $$\frac{{ - 1}}{2}$$ given Now, Required equation of the line = y - y 1 = m(x - x 1 ) = y - 3 = $$\frac{{ - 1}}{2}$$ (x - 2) x + 2y = 8

$$\eqalign{ & {\text{O is mid point of AB}} \cr & x = \frac{{5 + 0}}{2} = \frac{5}{2} \cr & y = \frac{{ - 3 + 2}}{2} = \frac{{ - 1}}{2} \cr & {\text{Slope of AB }}\left( {{m_1}} \right) \cr & = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \cr & = \frac{{2 - \left( { - 3} \right)}}{{0 - 5}} \cr & = \frac{5}{{ - 5}} \cr & = - 1 \cr} $$ Slope of (m 2 ) CD = 1 [lines are perpendicular to each other] m 1 × m 2 = -1 Equation of line $$\eqalign{ & \left( {y + \frac{1}{2}} \right) = 1\left( {x - \frac{5}{2}} \right) \cr & \Rightarrow 2\left( {x - y} \right) = 6 \cr & \Rightarrow x - y = 3 \cr} $$

(x - a) 2 + (y - b) 2 = r 2 Origin point = (0, 0) (x - 0) 2 + (y - 0) 2 = 6 2 x 2 + y 2 = 36 x 2 + y 2 - 36 = 0

$$\eqalign{ & x = \frac{{4 \times \frac{7}{2} + 1 \times 6}}{{4 + 1}} \Rightarrow 4 \cr & y = \frac{{4 \times 6 + 1 \times 1}}{{4 + 1}} \Rightarrow 5 \cr & {\text{A}}\left( {4,\,5} \right) \cr} $$

$$\eqalign{ & {\text{By distance formula, }} \cr & {\text{PQ}} = \sqrt {{{\left( {0 - 2} \right)}^2} + {{\left( {\frac{3}{2} - 0} \right)}^2}} \cr & = \sqrt {4 + \frac{9}{4}} \cr & = \frac{{\sqrt {25} }}{2} \cr & = \frac{5}{2} \cr & = 2.5{\text{ cm}} \cr} $$

$$\eqalign{ & \left( {y - {y_1}} \right) = \left( {\frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)\left( {x - {x_1}} \right) \cr & \Rightarrow \left( {y - 6} \right) = - \frac{1}{3} \times \left( x \right) \cr & \Rightarrow 3y - 18 = - x \cr & \Rightarrow x + 3y = 18 \cr} $$

$$\eqalign{ & \,\,\,\,4x - 2y = 10 \cr & \,\,\,\,4x + ky = 2 \cr & \underline {\, - \,\,\,\,\,\, - \,\,\,\,\,\,\,\,\, - \,\,\,\,} \cr & \left( { - k - 2} \right)y = 8 \cr & {\text{The quardinate of }}y{\text{ is}} = 4 \cr & \left( { - k - 2} \right) \times 4 = 8 \cr & - k - 2 = 2 \cr & k = - 4 \cr} $$

Equation of line perpendicular to line 3x - 2y = 6 is 2x + 3y + p = 0 As, the line passes through origin (0, 0) ∴ 2 × 0 + 3 × 0 + p = 0 ∴ p = 0 Now, the equation is 2x + 3y = 0 ∴ 2x = -3y x = $$\frac{{ - 3}}{2}$$ y By putting this value in equation 3x - 2y = 6 $$\eqalign{ & \Rightarrow 3\left( {\frac{{ - 3}}{2}} \right)y - 2y = 6 \cr & \Rightarrow \frac{{ - 9}}{2}y - 2y = 6 \cr & \Rightarrow \frac{{ - 13y}}{2} = 6 \cr & \Rightarrow y = - \frac{{12}}{{13}} \cr & \therefore x = \frac{{ - 3}}{2} \times \left( {\frac{{ - 12}}{{13}}} \right) \cr & x = \frac{{18}}{{13}} \cr & \therefore {\text{Co - ordinate of point}} \cr & {\text{M}} = \left[ {\frac{{18}}{{13}},\, - \frac{{12}}{{13}}} \right] \cr} $$

Slope of line y = mx + c is m y = 2x + c, slope (m 1 ) = 2 Lines are perpendicular to each other $${m_1} = \frac{{ - 1}}{{{m_2}}}$$ Slope of perpendicular line $$\left( {{m_2}} \right) = \frac{{ - 1}}{2}$$ on x axis y = 0 Equation of line which passes through (-5, 0) (y - y 1 ) = slope (m 2 )(x - x 1 ) y - 0 = $$\frac{{ - 1}}{2}$$ (x + 5) 2y = -x - 5 x + 2y = -5

$$\eqalign{ & 3x + 4y = 12 \cr & \Rightarrow \frac{{3x}}{{12}} + \frac{{4y}}{{12}} = 1 \cr & \Rightarrow \frac{x}{4} + \frac{y}{3} = 1 \cr & \therefore {\text{Length of intercept AB}} = \sqrt {{4^2} + {3^3}} \cr & = \sqrt {25} \cr & = 5{\text{ units}} \cr} $$

Slope of line PR $$\eqalign{ & \Rightarrow {m_1} = \frac{{0 - 2}}{{ - 2 - 4}} \cr & \Rightarrow {m_1} = \frac{{ - 2}}{{ - 6}} \cr & \Rightarrow {m_1} = \frac{1}{3} \cr} $$ ∴ Slope of line QS = -3 = m 2 {As it is perpendicular to PR} Coordinates of point O $$ \Rightarrow \left[ {\frac{{ - 2 + 4}}{2},\,\frac{{2 + 0}}{2}} \right] \Rightarrow \left( {1,\,1} \right)$$ ∴ Equation of line QS which passes through point O(1, 1) ⇒ y - 1 = m 2 (x - 1) ⇒ y - 1 = -3(x - 1) ⇒ y - 1 = -3x + 3 ⇒ 3x + y = 4

$$\eqalign{ & 2x + 3y = - 6 \cr & y = - \frac{2}{3}x - \frac{6}{3} \cr & \therefore {\text{Slope}} = - \frac{2}{3} \cr} $$ ⇒ Two lines are perpendicular if m 1 × m 2 = -1 Y intercept = 3 ⇒ y = mx + c ⇒ m 2 = $$\frac{3}{2}$$ The equation of line with slope $$\frac{3}{2}$$ is:- y - y 1 = m 2 (x - x 1 ) where, (x 1 y 1 ) = (0, 3) y - 3 = $$\frac{3}{2}$$(x - 0) 2y - 6 = 3x 3x - 2y = -6

Given, Equation of line x - 6y = 8 we write it as $$y = \frac{x}{6} - \frac{8}{6}$$ ∴ y = mx (where m is a slope) ∴ m 1 = $$\frac{1}{6}$$ If lines are perpendicular then product of their slopes is equal to -1 m 1 × m 2 = -1 ∴ $$\frac{1}{6}$$ × m 2 = -1 m 2 = - 6 ∴ Slope of perpendicular line = - 6 Perpendicular line which passes through the points (5, a) and (4, 3) $$\eqalign{ & \therefore \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = {m_2} \cr & \Rightarrow \frac{{3 - a}}{{4 - 5}} = - 6 \cr & \Rightarrow 3 - a = + 6 \times + 1 \cr & \therefore a = - 3 \cr} $$

Equation of line which intercepts x-axis and y-axis are given below:- $$\eqalign{ & \Rightarrow \frac{x}{a} + \frac{y}{b} = 1 \cr & {\text{where, }}a = \frac{3}{4},\,b = \frac{{ - 2}}{3}\,\,\left[ {{\text{given}}} \right] \cr & \Rightarrow \frac{x}{{\frac{3}{4}}} + \frac{y}{{\frac{{ - 2}}{3}}} = 1 \cr & \Rightarrow \frac{{4x}}{3} - \frac{{3y}}{2} = 1 \cr & \Rightarrow 8x - 9y = 6 \cr} $$

$$\eqalign{ & 3x + y = 5 \cr & \underline {2x - y = 5} \cr & 5x = 10 \cr & x = 2,\,y = - 1 \cr & \left( {\alpha ,\,\beta } \right) = \left( {2,\, - 1} \right) \cr & 3\alpha + \beta = 3 \times 2 - 1 = 5 \cr} $$

Slope (m 1 ) for the line which passes through the points (-3, 1) and (x, 5) $$ = \frac{{5 - 1}}{{x + 3}} = \frac{4}{{x + 3}}$$ Similarly, slope (m 2 ) for the line which passes through the points (-2, -1) and (6, 3) $$ = \frac{{3 + 1}}{{6 + 2}} = \frac{4}{8} = \frac{1}{2}$$ If two lines are parallel to each other. $$\eqalign{ & {\text{Then, }}{m_1} = {m_2} \cr & \Rightarrow \frac{4}{{x + 3}} = \frac{1}{2} \cr & \Rightarrow x + 3 = 8 \cr & \Rightarrow x = 5 \cr} $$

Given, $$\eqalign{ & m = \frac{{ - 2}}{3} \cr & m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \cr & \Rightarrow \frac{{ - 2}}{3} = \frac{{2 + 3}}{{5 - x}} \cr & \Rightarrow \frac{{ - 2}}{3} = \frac{5}{{5 - x}} \cr & \Rightarrow - 10 + 2x = 15 \cr & \Rightarrow 2x = 25 \cr & \Rightarrow x = 12.5 \cr} $$

The point in 4 th quadrant that is 6 unit away from x-axis and 7 unit away from y-axis is (7, -6).

By the midpoint formula, $$\left( {\frac{{{x_1} + {x_2}}}{2},\,\frac{{{y_1} + {y_2}}}{2}} \right)$$ Midpoint of line AC $$ = \left( {\frac{{7 + 1}}{2},\,\frac{{ - 8 + 4}}{2}} \right) = \left( {4,\, - 2} \right)$$ O(x 3 , y 3 ) = (4, -2) (M 1 ) Slope of line AC, $${M_1} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{4 - \left( { - 8} \right)}}{{1 - 7}} = - 2$$ If the lines are ⊥, then M 1 × M 2 = -1 - 2 × M 2 = -1 M 2 = $$\frac{1}{2}$$ ∴ Equation of line BD y - y 3 = M 2 (x - x 3 ) y - (-2) = $$\frac{1}{2}$$(x - 4) y + 2 = $$\frac{1}{2}$$(x - 4) x - 2y = 8

The equation whose graph passes through origin must satisfy point (0, 0) in the equation means if we put value of x & y equal to zero, the equation must be equal to zero. ∴ 2x + 3y = 0