# What is the equation of the line through the point (3, 4) and perpendicular to 7x + 5y – 10 =0?

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If a line is said to be perpendicular to a given line, its slope should be equal to the negative reciprocal of the given line.

The given line is 7x + 5y - 10 = 0. To get the slope of this given equation, we have to transform this equation in the slope-intercept form y = mx +b. The equation becomes,

5y = -7x + 10

y = (-7/5)x + 2

so the slope is -7/5.

So the slope of the new line perpendicular to the given line is 5/7 since it should be negative reciprocal, and it should pass the point (3,4)

Using point-slope form y - y1 = m(x - x1)

y - 4 = (5/7)(x - 3)

Multiply the whole equation by 7 it becomes,

7y - 28 = 5(x - 3)

7y - 28 = 5x - 15

So transposing all the terms in one side, the equation becomes

**5x - 7y +13 = 0.**

Given the point ( 3,4) and the line ( 7x + 5y -10 = 0

The line passes though the point (3,4) is given by:

y- 4 = m (x -3) where m is the slope.

Since the line is perpendicular to the line 7x+5y-10 = 0

Then the product of their slopes equals -1.

We will rewrite in the slope form:

7x+ 5y -10 = 0

==> 5y = -7x + 10

==> y= ( -7/5) + 2

Then, the slope = -7/5

==> -7/5 * m = -1

==> m = 5/7

==> y- 4 = (5/7)( x- 3)

==> y-4 = (5/7)x - 15/7

==> y= (5/7)x - 15/7 + 4

**==> y= (5/7)x + 13/7**

The equation 7x + 5y – 10 =0 can be written as 5y = -7x + 10 => y = (-7/5) x + 2.

Here the slope of the line is -7/5. The slope of a line perpendicular to this line is 5/7 as the product of the slope of two perpendicular lines is -1.

The line with a slope 5/7 and passing through (3, 4) is given as

y – 4 = (5/7)*(x – 3)

=> y – 4 = 5x / 7 – 15 /7

=> 7y – 28 = 5x – 15

=> 5x – 7y + 13 = 0

**The equation of the line passing through the point (3, 4) and perpendicular to 7x + 5y – 10 =0 is 5x – 7y + 13 = 0.**

Slope of a line

ax + by + c = 0 is given by:

Slope of line = -a/b

And slope of a line perpendicular to it is given by

Slope of perpendicular line = b/a

Thus the equation of the perpendicular line is:

bx - ay + c1 = 0

Substituting the given values of a and b in above line of equation the equation becomes:

5x - 7y + c1 = 0 ... (1)

o find the values of c1 for the perpendicular line passing through the point (3, 4) we substitute these coordinates in the above equation:

5*3 - 7*4 + c1 = 0

15 - 28 = - c1

c1 = 13

Substituting this value of c1 in equation (1) we get:

5x - 7y + 13 = 0

To find the the equation of the line through the point (3, 4) and perpendicular to 7x + 5y – 10 =0.

We know that the line perpendicular to ax+by+c = 0 is given by:

bx-ay +k = 0, where k is any constant to be determined by the given condition.

Therefore the line perpendicular to 7x+5y-10 = 0 is given by:

5x-7y+ k = 0....(1). Since this line also passes through the point (3,4), the coordinates of (3,4) should satisfy 5x-7y+k = 0.

So 5*3-7*4+k = 0.

15-28 +k = 0.

k = 28 -15 = 13.

Therefore substituting k= 1, in the equation at (1), we get:

5x-7y +13 = 0.

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