Practice Questions on Coordinate Geometry

Coordinate geometry is essential for anyone who wants to study mathematics, physics, engineering, or computer science.

Coordinate geometry is a branch of mathematics that uses algebraic equations to describe and analyze geometric shapes and their properties on a coordinate plane. In this article, we will look at some formulas along with a series of Practice Questions on Coordinate Geometry to help you sharpen your skills and deepen your understanding of this foundational topic

Important Formulas of Coordinate Geometry

• General Form of a Line : Ax + By + C = 0

• Slope Intercept Form of a Line : y = mx + c

• Point-Slope Form : y − y1= m(x − x1)

• The slope of a Line Using Coordinates : m = Δy/Δx = (y2 − y1)/(x2 − x1)

• The slope of a Line Using General Equation : m = −(A/B)

• Intercept-Intercept Form : x/a + y/b = 1

• Distance Formula : |P₁P₂| = √[(x2 − x1)² + (y2 − y1)²]

• For Parallel Lines : m1 = m2 ( m1, m2 are slope of the lines )

• For Perpendicular Lines : m₁m₂ = -1 ( m1, m2 are slope of the lines )

• Midpoint Formula : M (x, y) = [½(x1 + x2), ½(y1 + y2)]

• Angle Formula : tan θ = [(m1 – m2)/ (1 + m1m2)]

• Area of a Triangle Formula : Area = ½ |x1(y2−y3)+x2(y3–y1)+x3(y1–y2)|

• Distance from a Point to a Line : d = [|Ax₀ + By₀ + C| / √(A² + B²)]

Practice Questions on Coordinate Geometry - Solved

Question 1. Find the slope of the line passing through the points (2, 5) and (6, 9).
Solution:

To find the slope of the line passing through the points (2, 5) and (6, 9), use the following formula
m = (y2 - y1)/(x2 - x1)
m = 1.

So, the slope of the line passing through the points (2, 5) and (6, 9) is 1.

Question 2. Find the distance between the points (2, 3) and (5, 7) using the distance formula.
Solution:

The distance formula is :d = √[(x2 − x1)² + (y2 − y1)²]

Given points are (2, 3) and (5, 7).

Now,
d = √[(5 - 2)² + (7 - 3)²
d = √(3)² + (4)²
d = √9 + 16
d = √25
d = 5

So, the distance between the points (2, 3) and (5, 7) is 5.

Question 3. Find the midpoint of the line segment with endpoints (1, 2) and (5, 8).
Solution:

To find the midpoint use the following formula
M(x, y) = [(x1 + x2)/2, (y1 + y2)/2]

now,
M(x, y) = [(1+5)/2, (2+8)/2]
M(x, y) = [6/2, 10/2]
M(x, y) = (3, 5)

Hence, the middle point is (3, 5).

Question 4. Find the equation of a line passing through the point (2, 3) with slope 2 in slope-intercept form y = mx + c.
Solution:

The point-slope form equation of line is

y - y1 = m(x - x1)
Given point = (2, 3)
slope = m =2

Now,

y - 3 = 2(x - 2)
y - 3 = 2x - 4
y = 2x - 1

So, the equation of a line passing through the point (2, 3) with slope 2 is y = 2x - 1.

Question 5. Find the equation of a line which passes through the points (3, 4) and (5, 8) in the general form.
Solution:

To find the equation of a line in the general form, follow these steps:

Step 1: Find the slope (m):
m= (8-4)/(5-3)
m = 2

Step 2: Use the point slope form through (3, 4)
y - 4 = 2(x - 3)
y - 4 = 2x - 6
y = 2x - 2

Step 3: Now, convert to the general form
2x - y - 2 =0

Question 6. Find the equation of a line with x-intercept 4 and y-intercept 5 in intercept-intercept form x/a + y/b = 1.
Solution:

Given,
x-intercept = 4
y- intercept = 5

equation of line = x/a + y/b = 1.
x/4 + y/5 = 1
5x + 4y = 20.

So, the equation of a line with x-intercept and y-intercept is 5x + 4y = 20.

Question 7: Check if the points P(1,4), Q(3, 8), and R(5,12) are collinear. If they are, prove it using the slope formula.
Solution:

To check if the points are collinear, we need to verify if the slope between any two pairs of points is the same. If the slopes are equal, the points are collinear.

The formula for the slope between two points (x1,y1) and (x2,y2) is:

slope = \frac{y_2 - y_1}{x_2 - x_1}

First calculate the slope between P(1,4) and Q(3,8):

slope of PQ = (8 − 4)/(3 − 1) = 4/2 = 2

Now, calculate the slope between Q(3, 8) and R(5, 12):

slope of QR = (12 − 8)/(5 − 3) = 4/2 = 2

Since the slope PQ and QR is the same, we can say that the points P, Q, R are collinear.

Thus, the points P(1,4), Q(3,8), and R(5,12) are collinear.

Question 8 : Given the vertices of a triangle are A(2,3), B(5,7), and C(8,2), find the area of the triangle formed by these points using the formula for the area of a triangle in the coordinate plane.
Solution:

Area of the triangle formula: Area = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|

Where A(x1,y1), B(x2,y2), and C(x3,y3) are the coordinates of the vertices of the triangle.

Substituting the coordinates of A, B and C into the formula:

Area = \frac{1}{2} \left| 2(7 - 2) + 5(2 - 3) + 8(3 - 7) \right|
Area = \frac{1}{2} \left| 2(5) + 5(-1) + 8(-4) \right|
Area = \frac{1}{2} \left| 10 - 5 - 32 \right|
Area = \frac{1}{2} \left| -27 \right|
Area = \frac{1}{2} \times 27
Area =13.5

Thus, the area of the triangle is 13.5 square units.

Question 9. Find the area of the triangle formed by the vertices (4, 5), (10, 12) and (-3, 2).
Solution:

To find the area use the following formula Area = ½ |x1(y2−y3)+x2(y3–y1)+x3(y1–y2)|

so,
area of triangle = x1y2 - x1y3 + x2y3 - x2y1 + x3y1 - x3y2
area of triangle = 1/2 [4×12 - 4×2 + 10×2 - 10×(5) + (-3)×5 - (-3)×12]
area of triangle = 1/2[48 - 8 + 20 - 50 - 15 + 36]
area of triangle = 1/2[31]
area of triangle = 31/2

So, the area of triangle is 31/2.

Question 10. Find the distance from the point (2, 5) to the line 3x+4y−12=0.
Solution:

To find the distance from a point to the line , use this formula : d = [|Ax0 + By0 + C| / √(A² + B²)]

Now,
given line = (Ax + By + C = 0) = 3x + 4y -12 = 0

so, A = 3, B = 4 and C = -12
given point = (x0, y0) = (2, 5)

So, x0 = 2, y0 = 5
d = [|Ax0 + By0 + C| / √(A2 + B2)]
d = [|3 × 2 + 4 × 5 - 12| / √(3² + 4²)]
d = [|6 + 20 - 12| / √(9 + 16)]
d = [|14| / √(25)]
d = 14/5

So, the distance from a point (2, 5) to the line (3x + 4y - 12 = 0) is 14/5

Related Articles:

• Coordinate Geometry
• Practice Questions on Section Formula
• Practice Question on Distance Formula

Practice Questions on Coordinate Geometry - Unsolved

Q1. Find the equation of a line passing through the points (-1, 3) and (2, -5) in the general form Ax + By + C = 0.

Q2. Find the equation of a line passing through the point (3, -2) with slope -1 in slope-intercept form y = mx + c.

Q3. Find the equation of the line passing through the point (-4, 6) with slope 4 in point-slope form ? − ?1 = ?(? − ?1).

Q4. Find the slope of the line passing through the points (-3, 5) and (7, 2).

Q5. Find the slope of the line represented by the equation 2x − 3y + 6 = 0.

Q6. Find the equation of a line with x-intercept 3 and y-intercept -4 in intercept-intercept form x/a + y/b = 1.

Q7. Determine if the lines with equations 4x + 2y − 8 = 0 and 8x + 4y − 16 = 0 are parallel.

Q8. Determine if the lines with equations 5x + 2y − 10 = 0 and 2x − 5y + 15 = 0 are perpendicular.

Q9. Find the distance between the points (-2, 4) and (6, -1) using the distance formula.

Q10. Find the midpoint of the line segment with endpoints (-3, 7) and (5, -4).

Answer Key to Unsolved Questions

1. 8x + 3y - 1 = 0
2. y = -x + 1
3. y - 6 = 4(x + 4)
4. −3/10
5. 2/3
6. x/3−y/4=1
7. Parallel
8. Perpendicular
9. \sqrt{89}
10. (1,3/2)