Question

13. Write an equation of the line that passes through the points (-7, 6) and (3, -4 )in slope-

intercept form.

Answer 1

Firstly we need to find the gradient of give two points as follows;

M= y

Answer 2

Answer: y = -x - 1

Explanation:

- Consider a straight line passing through (x, y) from the origin (0, 0). That line with a positive gradient of m and meets at a point (0, c) [y-intercept]

- It has a general equation as below;

`{ \rm{y = mx + c}} \\`

- So, consider the line given in our question; Let's find its slope m first;

`{ \rm{slope = (y _(2) - y _(1) )/(x _(2) - x _(1) ) }} \\`

- From the points given in the question, (-7, 6) and (3, -4)

• x_1 is -7
• x_2 is 3
• y_1 is 6
• y_2 is -4

`{ \rm{m = ( - 4 - 6)/(3 - ( - 7)) }} \\ \\ { \rm{m = ( - 10)/(10) }} \\ \\ { \underline{ \rm{ \: m = - 1 \: }}}`

- Therefore, our equation so far is y = -x + c. Our line has a negative slope that means it slants from top to bottom, its origin is its y-intercept

- Consider point (3, -4);

`{ \rm{y = - x + c}} \\ { \rm{ - 4 = - 3 + c}} \\ { \rm{c = - 1}}`

- y-intercept is -1

hence equation is y = -x - 1

`{ \boxed{ \delta}}{ \underline{ \mathfrak{ \: \: beicker}}}`