View Full Version : Math Help Is Needed ASAP!!

tesha05

09-13-2005, 10:41 PM

Find an equation of the line that satisfies the given conditions.

Through (1,7); parallel to the line passing through (2,5) and (-2,1).

Through (-2,-11); perpepndicular to the line passing through (1,1) and ( 5,-1).

Find the area of the triangle formed by the coordinate axes and the line 2y + 3x - 6 = 0.

Through (1,7); parallel to the line passing through (2,5) and (-2,1).

use m = (y2 -y1)/(x2-x1)

m = (5-1)/ (2--2)

m = 4/2

m = 2

parallel to means it has the same gradient/slope

y=mx + b

sub in the point (1,7) and m= 2

solve for b and then rewrite the equation in the general form.

Through (-2,-11); perpepndicular to the line passing through (1,1) and

( 5,-1).

...do the same as the previous working to find the gradient "m"

However perpendicular lines have gradients that multiply together to make -1.

So m1 times m2 = -1 where m1 is the gradient from the pair of points and m2 is the gradient for your new line. Sub m2 and the point(-2,-11) into y=mx + b , solve for b and write the line equation.

Find the area of the triangle formed by the coordinate axes and the line 2y + 3x - 6 = 0.

Find the points where this line connects with the x and y axis.

x intercept occurs when y = 0

2(0) + 3x -6 =0

y intercept occurs when x=0

2y +3(0) -6 =0

solve for x and y in each of the above and write as a pair of points.

This gives you the length of 2 sides of the right angle triangle.

Area = (1/2) x side1 x side2

over to you...............

Powered by vBulletin® Version 4.2.2 Copyright © 2015 vBulletin Solutions, Inc. All rights reserved.