Practice Problems with Solutions

Q1. 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

Q2. 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.

Q3. 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 = (9 – 5)/(6 – 2)

m = 1.

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

Q4. 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.

Q5. 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.

Q6. Determine if the lines with equations 2x+3y−5=0 and 4x+6y−10=0 are parallel.

Solution:

To check whether the given pair of lines are parallel, we need to check their slope

if m1 = m2, then they are parallel.

Now, to calculate slope of first line 2x + 3y -5 = 0

m1 = -(a/b)

m1 = -2/3

Now, we calculate slope of second line 4x + 6y – 10 = 0

m2 = -(a/b)

m2 = -(4/6)

m2 = -2/3

Here, m1 = m2

So, the lines 2x + 3y – 5 = 0 and 4x + 6y – 10 = 0 are parallel lines.

Q7. Determine if the lines with equations 3x+4y−7=0 and 4x−3y−5=0.

Solution:

To check whether the given pair of lines are perpendicular, we need to check their slope

if m1.m2 = -1, then they are perpendicular.

Now, to calculate slope of first line 3x + 4y – 7 = 0

m1 = -(a/b)

m1 = -3/4

Now, we calculate slope of second line 4x – 3y – 5 = 0

m2 = -(a/b)

m2 = -(4/-3)

m2 = 4/3

now, m1.m2 = (-3/4).(4/3)

= -1

So, the lines 2x + 3y – 5 = 0 and 4x + 6y – 10 = 0 are perpendicular lines.

Q8. 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).

Q9. 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[71]

area of triangle = 71/2

So, the area of triangle is 71/2.

Q10. 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

In this article, we will learn about one interesting topic which is covered in class 9 and class 10 mathematics. We will look at some formulas and problems of Coordinate Geometry.