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September 19, 2023
This page deals with different types of Coordinate geometry Questions and Answers which can be asked in placement exams or any other competitive exa ms which has included Coordinate Geometry in its Syllabus.
Example 1:
Find the equation of straight line passing through (2, 3) and perpendicular to the line 3x + 2y + 4 = 0
Options:
a. y=5/3x- 2b. 3Y=2x+5c. 3Y=5x-2d. None of these
Solution:
The given line is 3x + 2y + 4 = 0 or y = -3x / 2 – 2Any line perpendicular to it will have slope = 2 / 3Thus equation of line through (2, 3) and slope 2 / 3 is(y – 3) =\frac{2}{3(x – 2)}3y – 9 = 2x – 43y – 2x – 5 = 0.
Correct Option is b.
Example 2:
Find the coordinate of the point which will divide the line joining the point (2,4) and (7,9) internally in the ratio 1:2?
a. (5/3 , 1/3)b. (3/8 , 3/11)c. (8/3 , 11/3)d. (11/3 , 17/3)
Answer:
The internal division will use the formula \frac{(mx_2 + nx_1)}{(m + n)}y = \frac{(my_2 + ny_1)}{(m + n)}.So, the point becomes (11/3, 17/3).
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1. Given two points A(-2,0) and B(0, 4), M is a point with coordinates (x, x), x ≥ 0. P divides the joining of A & B in the ratio 2: 1. C & D are the midpoints of BM and AM respectively. Area of the ∆AMB is minimum if the coordinates of M are
(0,0)
(1,1)
(2,2)
Area of [llatex]\Delta AMB = \frac{1}{2}\begin{vmatrix} x & x& 1\\ -2 & 0& 1\\ 0 & 4& 1 \end{vmatrix}[/latex] \left | \frac{1}{2}(-4x + 2x - 8) \right | = \left | -(x + 4) \right | which minimum for x = 0 and thus the co-ordinates of M are (0,0)
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2. If the centroid of a triangle formed by (7, x), (y, –6), and (9, 10) is (6, 3), then the values of x and y are respectively.
5,2
2,5
1,0
0,0
Centroid = \left ( \frac{x_1 + x_2 + x_3}{3} \right ), \left ( \frac{y_1 + y_2 + y_3}{3} \right ) \Rightarrow (6,3) = \left ( \frac{7 + y + 9}{3} \right ), \left ( \frac{x - 6 + 10}{3} \right ) \Rightarrow 6 = \frac{7 + y + 9}{3} \text{ and }3 = \frac{x - 6 + 10}{3}
y = 2 and x = 5
3. (0, 0, 0) (a, 0, 0), (0, b, 0) and (0, 0, c) are four distinct points. What are the coordinates of the point which is equidistant from the four points?
((a+b+c)/3, (a+b+c)/3, (a+b+c)/3)
(a, b, c)
(a/3, b/3, c/3)
(a/2, b/2, c/2)
OP = \sqrt{\left ( \frac{a}{2} - 0 \right )^{2} + \left ( \frac{b}{2} - 0 \right )^{2} + \left ( \frac{c}{2} - 0 \right )^{2} } = \sqrt{\frac{a^2 + b^2 + c^2}{4}} OQ = \sqrt{\left ( \frac{a}{2} - a \right )^{2} + \left ( \frac{b}{2} - 0 \right )^{2} + \left ( \frac{c}{2} - 0 \right )^{2} } = \sqrt{\frac{a^2 + b^2 + c^2}{4}} OR = \sqrt{\left ( \frac{a}{2} - 0 \right )^{2} + \left ( \frac{b}{2} - b \right )^{2} + \left ( \frac{c}{2} - 0 \right )^{2} } = \sqrt{\frac{a^2 + b^2 + c^2}{4}} OS = \sqrt{\left ( \frac{a}{2} - 0 \right )^{2} + \left ( \frac{b}{2} - b \right )^{2} + \left ( \frac{c}{2} - c \right )^{2} } = \sqrt{\frac{a^2 + b^2 + c^2}{4}}
4. What is the ratio in which the point C(-2/7,-20/7) divides the line joining the points A(–2, –2) and B(2, –4)?
1 : 3
3:4
1:2
2:3
C=\frac{2K-2}{K+1},\frac{-4K+2}{K+1} [Latex]\frac{2K-2}{K+1}=\frac{-2}{7}\[/latex] C(\frac{-2}{7},\frac{-20}{7}) \frac{2(K-1)}{K+1} =\frac{-2}{7} 7K-7 = -K-1 8K =6=K=[Latex]\frac{3}{4}[/latex] K:1 = \frac{3}{4}:1= 3:4
5. The co-ordinates of incentre of ∆ ABC with vertices A(0, 6), B(8, 12), and C(8, 0) is.
(5, 6)
(–4, 3)
(8, 11)
(16/3,0)
a = BC = \sqrt{0^{2} + (12 - 0)^{2}} = 12 b = AC = \sqrt{(0 - 8)^2 + (6 - 0)^{2}} = 10 c = AB = \sqrt{8^2 + 6^2} = 10 Incenter is \left ( \frac{ax_1 + bx_2 + cx_3}{a + b + c} \right ),\left ( \frac{ay_1 + by_2 + cy_3}{a + b + c} \right ) i.e \left ( \frac{12\times 0 + 10\times 8 + 10\times 8}{12 + 10 + 10} \right ), \left ( \frac{12\times 6 + 10\times 12 + 10\times 0}{12 + 10 + 10} \right ) \left ( \frac{160}{32}, \frac{192}{32} \right ) = (5,6)
6. If t₁ ≠ t₂ and the points A (a, 0), B (at₁², 2at₁) and C (at₂², 2at₂) are collinear, then t₁ t₂ is equal to
1
-1
2
-2
\Delta = \frac{1}{2}\begin{vmatrix}<br /> a & 0& 1\\<br /> at_1^2& 2at_1& 1\\<br /> at_2^2 & 2at_2& 1<br /> \end{vmatrix} = \frac{1}{2}\times (2a)\times a\times \begin{vmatrix}<br /> 1 & 0& 1\\<br /> t_1^2 & t_1& 1\\<br /> t_2^2 & t_2& 1<br /> \end{vmatrix} \therefore \Delta = 0 \Rightarrow (t_1 - t_2)+(t_1^2t_2 - t_2^2t_1)= 0 \Rightarrow (t_1 - t_2) + t_1t_2(t_1 - t_2) = 0 \Rightarrow (t_1 - t_2)(1 + t_1t_2)= 0 \Rightarrow t_1t_2 = -1
7. The area of quadrilateral ABCD whose vertices in order are A(1, 1), B(7, –3), C(12, 2) and D(7, 21) is
132 sq units
66 sq units
124 sq units
86.5 sq units
Area of quadrilateral ABCD: =\frac{1}{2}\begin{vmatrix}<br /> x_1 - x_3 & y_1 - y_2\\<br /> x_2 - x_4 & y_2 - y_3<br /> \end{vmatrix} =\frac{1}{2}\begin{vmatrix}<br /> 1 - 12 & 1 - 2\\<br /> 7 - 7 & -3 - 21<br /> \end{vmatrix} =\frac{1}{2}\begin{vmatrix}<br /> -11 & -1\\<br /> 0 & -24<br /> \end{vmatrix} \frac{1}{2}(264 - 0) = 132sq.m
8. The vertices of a triangle ABC are A (2, 3, 1), B (–2, 2, 0), and C(0, 1, –1). What is the magnitude of the line joining mid points of the sides AC and BC?
1/√2 unit
1 unit
3/√2 unit
2 unit
Mid Point of A and C \left ( \frac{2+0}{2}, \frac{3 + 1}{2}, \frac{1 - 1}{2} \right ) = (1, 2, 0) Mid Point of B and C \left ( \frac{-2+0}{2}, \frac{2 + 1}{2}, \frac{0 - 1}{2} \right ) = \left ( -1, \frac{3}{2}, \frac{-1}{2} \right ) Magnitude =\sqrt{(1 + 1)^2 + \left ( 2 - \frac{3}{2} \right )^2 + \left ( \frac{1}{2} \right )^2} =\sqrt{4 + \left ( \frac{2}{4} \right )+ \left ( \frac{1}{4} \right )}=\frac{3}{\sqrt{2}}
9. What is the reflection of the point (1.6, 5) in the line y = 1.4?
(1.6, 2.2)
(1.6, –1.4)
(1.6, –3.6)
(1.6, –2.2)
10. A Rod is placed in two-dimensional Co-ordinate plane and co-ordinate of its ends are A=(6, – 4) and B=(0, 8). A carpenter wants to cut this road in 5:1. What are the co-ordinates of that point where he should cut?
(–1, 6)
(1, 6)
(–1, –6)
(1, –6)
X = \frac{5\times 0 + 1\times 6}{6} = 1 Y = Y = \frac{5\times 8 + 1\times -6}{6} = 6 C = (X,Y) = (1,6)
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