22.1k views
4 votes
Find the slope of the line joining the points (-1,5) and (6,-2). Also find the equation of that straight line. Find the intercepts on the x- and y axis

1 Answer

4 votes

The slope of the line joining the points (-1,5) and (6,-2) is -1

x + y = 4 is the equation of line

x intercept is (-4, 0)

y intercept is (0, 4)

Solution:

Given that,

Points are (-1,5) and (6,-2)

The slope of line is given as:


m = (y_2-y_1)/(x_2-x_1)

From given,


(x_1, y_1) = (-1, 5)\\\\(x_2, y_2) = (6, -2)

Substituting the values we get,


m = (-2-5)/(6+1)\\\\m = (-7)/(7)\\\\m = -1

Thus slope of line is -1

The equation of line in slope intercept form is given as:

y = mx + c -------- eqn 1

Where,

m is the slope

c is the y intercept

Substitute m = -1 and (x, y) = (-1, 5) in eqn 1

5 = -1(-1) + c

5 = 1 + c

c = 4

Substitute m = -1 and c = 4 in eqn 1

y = -x + 4

In standard form,

x + y = 4 is the equation of line

Find x intercept:

Substitute y = 0

x + 0 = 4

x = -4

Thus x intercept is (-4, 0)

Find y intercept:

Substitute x = 0

0 + y = 4

y = 4

Thus y intercept is (0, 4)

User MikeP
by
8.6k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories