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tesha05
09-13-2005, 11:41 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.

09-13-2005, 11:47 PM
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.

09-13-2005, 11:50 PM
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.

09-13-2005, 11:56 PM
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...............