What is the equation of the line perpendicular to 2x - 3y = 13 that passes through the point (-6, 5)?

2y + 3x = -8.

Step-by-step explanation:

Given our first equation 2x-3y =13, we rewrite in the form of equation of a line, i.e y = mx + c

Where m is the slope of the line and c is the intersection on y axis.

By rewriting, we have...

3y= 2x-13

y= 2x/3 - 13/3.

m= 2/3, c= -13/3.

For two perpendicular lines, the product of their gradients is -1

i.e

m¹ * m² = -1

Hence m² = -1 * 3/2

m² = -3/2.

Hence, to get the equation of the line passing through the point (-6,5) we first find the intersection when x=0 given as

[Y-c]/[x-0] = -3/2

2y - 2c = -3x

2(5) - 2c =- 3(-6)

10-18= 2c

C = -8/2

C= -4

Then the equation is given as

y = mx + c

y = [-3/2]x - 4

2y= -3x - 8

2y + 3x = -8.

70

step-by-step explanation:

2*5=10

4*15=60

60+10=

70