Coordinate Geometry — Part 1 MCQs with Explanations

$$\eqalign{ & \frac{{K\,\,\,\,\,:\,\,\,\,\,1}}{{S\left( { - 4,\, - 1} \right)\,\,\,\,\,\,\,\,\,\,T\left( {x,\,0} \right)\,\,\,\,\,\,\,\,\,\,U\left( {1,\,4} \right)}} \cr & \Rightarrow x = \frac{{k{x_2} + {x_1}}}{{k + 1}},\,\,y = \frac{{k{y_2} + {y_1}}}{{k + 1}} \cr & \Rightarrow 0 = \frac{{4k - 1}}{{k + 1}} \Rightarrow \boxed{k = \frac{1}{4}} \cr & {\text{Ratio }}\frac{1}{4}:1 = \boxed{1:4} \cr} $$

$$\eqalign{ & {\text{Now,}} \cr & \Rightarrow 1 = \frac{{2y - 9}}{5} \cr & \Rightarrow 5 = 2y - 9 \cr & \Rightarrow y = 7 \cr} $$

$$\eqalign{ & 7x + 11y = 3\,\,\,\,*1 \cr & \underline {\,8x + y = 15\,} \,\,\,\,\,\underline {\,*11\,} \cr & \,\,7x + 11y = 3 \cr & \,\,88x + 11y = 165 \cr & \underline {\, - \,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\,\,} \cr & - 81x = - 162 \cr & x = 2 \cr & y = 15 - 8 \times 2 \cr & y = - 1 \cr} $$ Put value of x and y in options only option 'D' satisfy in this values 3x + 5y = 1 3 × 2 + 5 × (-1) = 1 6 - 5 = 1 = 1 [satisfied]

x = k, where k is a constant The graph of the equation will be parallel to y-axis.

$$\eqalign{ & 4x + 5y = 40 \cr & \Rightarrow \frac{x}{{10}} + \frac{y}{8} = 1 \cr} $$ Intersection point of 4x + 5y = 40 and y = 4 will be, 4x + 5 × 4 = 40 4x = 20 x = 5 ∴ Intersecting point is (5, 4) Area of ΔABC = $$\frac{1}{2}$$ × (10 - 5) × 4 = 10 sq. units Area of ΔCDE = $$\frac{1}{2}$$ × 5 × (8 - 4) = 10 sq. units ∴ Area bounded by the graph = 10 + 10 = 20 sq units.

Since ordinate is four times its abscissa, equation of line will be y = 4x

$$\eqalign{ & \,\,x + y = 3 \cr & \underline {\,2x + 5y = 12\,} \cr & \,\,\,\,\,\,\,\,\,\,y = 2 \cr} $$ Area of ΔABC = $$\frac{1}{2}$$ × 3 × 2 = 3

Equation of straight line parallel to x-axis and also 3 units below x-axis is y = -3

(x - a) 2 + (y - b) 2 = r 2 (x - 1) 2 + [y - (-2)] 2 = 4 2 x 2 - 1 - 2x + y 2 + 4 + 4y = 16 x 2 + y 2 - 2x + 4y = 11

ax + by + c = 0 Where c = 0, a ≠ 0 & b ≠ 0 ∴ ax + by = 0 Since ax + by = 0, therefore x = 0 & y = 0 Therefore, the straight line passes through (0, 0)

At y-axis, x = 0 0 - 7y = -42 y = 6 (α, β) = (0, 6) At x-axis, y = 0 6x + 0 = 15 x = $$\frac{5}{2}$$ (γ, δ) = $$\left( {\frac{5}{2},\,0} \right)$$ α + β + γ + δ = 0 + 6 + $$\frac{5}{2}$$ + 0 = $$\frac{{17}}{2}$$

$$\eqalign{ & {m_1}\left( {{\text{slope}}} \right) = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \cr & \Rightarrow \frac{{ - 4 - \left( { - 1} \right)}}{{4 - 5}} \cr & \Rightarrow \frac{{ - 3}}{{ - 1}} = \left( 3 \right) \cr & \therefore {m_1} = {m_2}\left( {{\text{In parallel condition}}} \right) \cr & \therefore {m_2} = 3 \cr} $$

$$\eqalign{ & {\text{Area of }}\Delta ABC \cr & = \frac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right] \cr & = \frac{1}{2}\left[ {4\left( {2 - 14} \right) + 8\left( {14 - 2} \right) + 8\left( {2 - 2} \right)} \right] \cr & = \frac{1}{2}\left[ {4 \times \left( { - 12} \right) + 8\left( {12} \right) + 8\left( 0 \right)} \right] \cr & = \frac{1}{2}\left[ { - 48 + 96} \right] \cr & = \frac{1}{2} \times 48 \cr & = 24{\text{ sq}}{\text{. units}} \cr & {\text{Area of }}\Delta ACD \cr & = \frac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right] \cr & = \frac{1}{2}\left[ {4\left( {14 - 10} \right) + 8\left( {10 - 2} \right) + 4\left( {2 - 14} \right)} \right] \cr & = \frac{1}{2}\left[ {4 \times 4 + 8 \times 8 + 4 \times \left( { - 12} \right)} \right] \cr & = \frac{1}{2}\left[ {16 + 64 - 48} \right] \cr & = \frac{1}{2} \times 32 \cr & = 16{\text{ sq}}{\text{. units}} \cr & {\text{Hence Area of Quadrilateral }}ABCD \cr & = {\text{Area of }}\left( {\Delta ABC + \Delta ACD} \right) \cr & = 24 + 16 \cr & = 40{\text{ sq}}{\text{. units}} \cr} $$

$$\eqalign{ & 3x - 6y = 4 \cr & \Rightarrow 6y = 3x - 4 \cr & \Rightarrow y = \frac{{3x}}{6} - \frac{4}{6} \cr & \Rightarrow y = \frac{1}{2}x - \frac{2}{3} \cr & y = {m_1}x + c \cr & {\text{If lines are parallel then their slopes are equal}} \cr & \therefore {m_1} = {m_2} = \frac{1}{2} \cr} $$

$$\eqalign{ & 2x + 5y = 12 \cr & x{\text{ - axis}}:y = 0\,;\,x = 6 \cr & x + y = 3 \cr & x{\text{ - axis}}:y = 0\,;\,x = 3 \cr & 2x + 5y = 12\,\,\,\,\,*1 \cr & \underline {\,x + y = 3\,\,\,\,\,\,\,\,\,\,\,\,*2\,} \cr & \,\,\,\,\,2x + 5y = 12 \cr & \,\,\,\,\,2x + 2y = 6 \cr & \underline {\,\, - \,\,\,\,\,\, - \,\,\,\,\,\,\,\, - \,\,\,\,} \cr & \,\,\,\,\,3y = 6 \cr & \,\,\,\,\,\,y = 2 \cr & \,\,\,\,\,\,x = 3 - 2 = 1 \cr} $$ $$\eqalign{ & {\text{Area of traingle}} = \frac{1}{2} \times {\text{base}} \times {\text{height}} \cr & = \frac{1}{2} \times 3 \times 2 \cr & = 3 \cr} $$

Since ordinate = abscissa ∴ k = 2k - 1 k = 1

Reflection of point (4, -6) in origin is (-4, 6)

Go through option, Points (4, 3) and (-2, 1) will satisfy the equation from option B, x 2 + y 2 - 2x - 4y = 5

$$\eqalign{ & 5x - 2y = - 1\,.\,.\,.\,.\,.\,.\,.\,\left( {\text{i}} \right) \times 2 \cr & 4y - 3x = - 5\,.\,.\,.\,.\,.\,.\,.\,\left( {{\text{ii}}} \right) \cr & 10x - 4y = - 2 \cr & \underline {\, - 3x + 4y = - 5\,} \cr & 7x = - 7 \cr & x = - 1,\,\,y = - 2 \cr & 2\alpha - 3\beta \cr & = 2\left( { - 1} \right) - 3\left( { - 2} \right) \cr & = - 2 + 6 \cr & = 4 \cr} $$

Area of Δ formed by the graph x - y = 0, x + y = 2 and x-axis Intersection point of lines x - y = 0 and x + y = 2 will be (1, 1) ∴ Area bounded by lines x - y = 0, x - y = 2 and x-axis = $$\frac{1}{2}$$ × 2 × 1 = 1 sq unit

Point (a, b) = (3, -5) Slope = 2 The general equation of a straight line (y - b) = m(x - a) (y + 5) = 2(x - 3) y + 5 = 2x - 6 2x - y - 11 = 0

25x + 75y = 225 . . . . . . (i) x = 9 . . . . . . (ii) For meeting point, solving the equation (i) & (ii) ⇒ 25 × 9 + 75y = 225 ⇒ 75y = 225 - 225 ⇒ y = 0 ∴ Meeting point = (9, 0)

In given rectangle Let $$\square $$ ABCD Angle of rectangle = 90° AB ⊥ DA ∴ Slope $$\left[ {{{\text{m}}_1} = \frac{{ - 1}}{{{{\text{m}}_2}}}} \right]$$ (when lines are perpendicular to each other) Slope of side DA (m 1 ) = $$\frac{5}{3}$$ Slope of side AB (m 2 ) = $$\frac{{ - 3}}{5}$$

Co-ordinates of mid point (D) which lies on the line BC = $$\left( {\frac{{3 + 5}}{2},\,\frac{{0 - 2}}{2}} \right) = \left( {4,\, - 1} \right)$$ Now, Co-ordinates of line AD = A(2, -4), D(4, -1) ∴ Equation of the line which passes through the two point (x 1 , y 1 ), & (x 2 , y 2 ) ⇒ y - y 1 = m(x - x 1 ) ∴ required equation of the line = y + 4 = m(x - 2) $$\eqalign{ & \left[ {\because m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right] \cr & \therefore m = \frac{{ - 1 + 4}}{{4 - 2}} = \frac{3}{2} \cr & \Rightarrow y + 4 = \frac{3}{2}\left[ {x - 2} \right] \cr & \Rightarrow 2y + 8 = 3x - 6 \cr & \Rightarrow 3x - 2y = 14 \cr} $$

$$\eqalign{ & 9x - 12y = 108 \cr & \Rightarrow \frac{{9x}}{{108}} - \frac{{12y}}{{108}} = 1 \cr & \Rightarrow \frac{x}{{12}} - \frac{y}{9} = 1 \cr} $$ $$\eqalign{ & {\text{Length of Intercept}} = \sqrt {{{12}^2} + {9^2}} \cr & = \sqrt {144 + 81} \cr & = \sqrt {225} \cr & = 15{\text{ units}} \cr} $$

Standard equation of the line y = mx + c ∴ m = tanθ = tan45° (θ = 45°) Given m = 1 ∴ c = $$ - \frac{3}{4}$$ New required equation of the line ⇒ y = mx + c ⇒ y = 1 × x $$ - \frac{3}{4}$$ ⇒ 4y = 4x - 3 ⇒ 4x - 4y = 3

$$\eqalign{ & {\text{Given,}} \cr & 4x - y = 1 \cr & \therefore y = 4x - 1 \cr & 5x - ky = 2 \cr & \therefore y = \frac{{5x}}{k} - \frac{2}{k} \cr & y = {m_1}x + C \cr & {m_1} = \frac{5}{k} \cr} $$ If the lines are ⊥ then the product of their slope is -1 \[\begin{array}{l} {m_1} \times {m_2} = - 1\\ 4 \times \frac{5}{k} = - 1\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l} {m_1} = 4\\ {m_2} = \frac{5}{k} \end{array} \right.\\ \therefore k = - 20 \end{array}\]

2x - 3y = 6 Cut Y-axis at x = 0 ∴ 0 - 3y = 6 $$\boxed{y = - 2}$$ Hence, point will be (0, -2)

Equation of given line $$\eqalign{ & \Rightarrow 20x + 5y = 3 \cr & \Rightarrow 5y = 3 - 20x \cr & \Rightarrow y = - 4x + \frac{3}{5} \cr} $$ Slope of line, m 1 = -4 If two are lines are ⊥ then the product of their slope = -1 $$\eqalign{ & {m_1} \times {m_2} = - 1 \cr & - 4 \times {m_2} = - 1 \cr & {m_2} = \frac{1}{4} \cr} $$ Lines passing through the points (-2, 5) and (6, b) Therefore, $$\eqalign{ & {\text{Slope}} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{b - 5}}{{6 + 2}} = \frac{{b - 5}}{8} \cr & {\text{According to the question,}} \cr & \frac{{b - 5}}{8} = \frac{1}{4} \cr & b - 5 = 2 \cr & \therefore b = 7 \cr} $$

$$\eqalign{ & 2x - 3y = 11 \cr & 3y = 2x - 11 \cr & y = \frac{2}{3}x - \frac{{11}}{3} \cr & \left( {y = mx + c,{\text{ where }}m{\text{ is slope}}} \right) \cr & {\text{Slope}} = {m_1} = \frac{2}{3} \cr & 3x + ky = - 4 \cr & ky = - 3x - 4 \cr & y = - \frac{3}{k}x - \frac{4}{k} \cr & {m_2} = - \frac{3}{k} \cr} $$ Relation between slope of perpendicular lines $$\eqalign{ & {m_1}{m_2} = - 1 \cr & \Rightarrow \left( {\frac{2}{3}} \right) \times \left( { - \frac{3}{k}} \right) = 1 \cr & \Rightarrow k = 2 \cr} $$

$$\eqalign{ & 3x + 4y - 24 = 0 \cr & \Rightarrow 3x + 4y = 24 \cr & \Rightarrow \frac{{3x}}{{24}} + \frac{{4y}}{{24}} = 1 \cr & \Rightarrow \frac{x}{8} + \frac{y}{6} = 1 \cr & {\text{Area of }}\Delta OAB = \frac{1}{2} \times 6 \times 8 = 24{\text{ sq}}{\text{. units}} \cr & {\text{And,}} \cr & x + y + 4 = 0 \cr & \Rightarrow x + y = - 4 \cr & \Rightarrow \frac{x}{{\left( { - 4} \right)}} + \frac{y}{{\left( { - 4} \right)}} = 1 \cr} $$ $$\eqalign{ & {\text{Area of }}\Delta OCD = \frac{1}{2} \times 4 \times 4 = 8{\text{ sq}}{\text{. units}} \cr & \therefore {\text{Area of }}\Delta OCD = \frac{1}{3}{\text{Area of }}\Delta OAB \cr} $$

Distance formula between two points Distance = $$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $$ ⇒ 13 = $$\sqrt {{{\left( {k - 4} \right)}^2} + {{\left( { - 4 - 8} \right)}^2}} $$ ⇒ 169 = k 2 + 16 - 8k + 144 ⇒ k 2 - 8k - 9 = 0 ⇒ k 2 - 9k + k - 9 = 0 ⇒ k(k - 9) + 1(k - 9) = 0 ⇒ (k + 1)(k - 9) = 0 ⇒ k = -1 and 9 ∴ k = -1 (According to options)

$$\eqalign{ & {\text{Slope}} \Rightarrow m = - 1 \cr & m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \cr & - 1 = \frac{{5 - \left( { - 1} \right)}}{{x - 2}} \cr & - 1\left( {x - 2} \right) = 6 \cr & - x + 2 = 6 \cr & x = 2 - 6 \cr & x = - 4 \cr} $$

$$\eqalign{ & 17x + my + 102 = 0 \cr & 23x + 299y + 138 = 0 \cr & {\text{Infinite solution}} \cr & \frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \cr & \frac{{17}}{{23}} = \frac{m}{{299}} \cr & m = \frac{{17 \times 299}}{{23}} = 221 \cr} $$

$$\eqalign{ & {\text{For coincident lines}} \cr & \frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}} \cr & \therefore \frac{k}{3} = \frac{2}{1} = \frac{2}{1} \cr & {\text{Hence, }}k = 3 \times 2 \cr & k = 6 \cr} $$

$$\eqalign{ & {\text{At }}x{\text{ - axis}},\,y = 0 \cr & 8x + 3y = 24 \cr & x = 3 \cr & 2x + 8 = y \cr & x = - 4 \cr & \,\,\,8x + 3y = 24 \to \left( {\text{i}} \right) \cr & \,\,\,8x - 4y = 32 \to \left( {{\text{ii}}} \right) \cr & \underline {\, - \,\,\,\,\, + \,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,} \cr & 7y = 56 \cr & y = 8 \cr & {\text{Area}} = \frac{1}{2} \times {\text{base}} \times {\text{height}} \cr & = \frac{1}{2} \times 7 \times 8 \cr & = 28 \cr} $$

Slope (m 1 ) of line which passes through two points (-3, 4) and (0, 3) $$\eqalign{ & {m_1} = \left( {\frac{{3 - 4}}{{0 + 3}}} \right)\,\,\,\,\,\left[ {\because m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right] \cr & \Rightarrow {m_1} = \frac{{ - 1}}{3} \cr} $$ Similarly, slope (m 2 ) of line which passes through the two points (5, 7) and (4, x) $${m_2} = \frac{{x - 7}}{{4 - 5}} = - \left( {x - 7} \right)$$ ∵ These lines perpendicular to each other, $$\eqalign{ & \therefore {m_1} \times {m_2} = - 1 \cr & \frac{{ - 1}}{3} \times \left[ { - \left( {x - 7} \right)} \right] = - 1 \cr & x - 7 = - 3 \cr & x = 4 \cr} $$

Given that (2, 0) is solution of equation 2x + 3y = k ∴ On putting value of x & y in the above equation, 2 × 2 + 3 × 0 = k k = 4

$$\eqalign{ & 59x + 14y - 112 = 0 \cr & \Rightarrow 59x + 14y = 112 \cr & \Rightarrow \frac{{59x}}{{112}} + \frac{{14y}}{{112}} = 1 \cr & \Rightarrow \frac{x}{{\frac{{112}}{{59}}}} + \frac{y}{8} = 1 \cr & \therefore y{\text{ - intercept of the line is }}8 \cr} $$

$$\eqalign{ & 2x + 3y = 11{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{i}} \right) \cr & x - 2y = - 12{\text{ }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\text{. }}{\left( {{\text{ii}}} \right)_{ \times 2}} \cr & \,2x - 4y = - 24 \cr & \underline {\,2x + 3y = 11\,} \cr & \,\,\,\,\,\,\,\,\,\,\,\,7y = 35 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,y = 5 \cr & {\text{From equation}}\left( {\text{i}} \right) \cr & 2x + 15 = 11 \cr & x = - 2 \cr & \left( {{x_1},\,{y_1}} \right) = \left( { - 2,\,5} \right) \cr & {\text{At }}x{\text{ - axis}},\,y = 0 \cr & x - 2 \times 0 = - 12 \cr & x = - 12 \cr & \left( {{x_2},\,{y_2}} \right) = \left( { - 12,\,0} \right) \cr & {x_1} - {x_2} + {y_1} + {y_2} = - 2 + 12 + 5 + 0 = 15 \cr} $$