b. b= -3c. x+(2/5)y=6/5x-(a/3)y=4/3Now two lines to be parallel, -(a/3)=2/5hence,......
b = -3 <------ slope
y = -3x - 3
This problem I can't figure out. Please help: Find a so that the two lines given are parallel: 5x + 2y = 6 and 3x - ay = 4
solve for y=
y = -5/2x + 3
y = 3/a - 4
set the slopes equal to each other and solve for a
-5/2 = 3/a * 2 on each side
-5 = 3/a * 2 * a on each side
-5a = 3*2
-5a = 6 /-5 on each side
a = -6/5
y = 3/(-6/5)x + 4/(6/5) <------
a. Are those points collinear - To do that you have to first find out the equation of the line and then put the value of x and y from the third point to see if it satisfies the equation.
Let us take points (2,1) and (3.3). X1 = 2, Y1 = 1, X2 = 3, Y2 = 3
Slope = (y2 - y1)/(x2 - x1)
= (3 - 1)/(3 - 1)
= 2
So the line's equation is Y = 2x + C. We need to find out C the Y intercept.
Put the value of first point in it.
1 = 2 x 2 + C
So C = -3
Verify by putting the value of second point into the equation Y = 2x -3
3 = 2 x 3 - 3
So the equation is correct.
Now put the value of X from third point and see if the equation yields the value of Y that is the same as that of the point.
Y = 2 x 4 - 3
Y = 8 - 3
Y = 5
That is same as the Y of the point.
So the three points are collinear.
b. To find the answer you need to find the slope of the lines. If slope is equal, they are parallel, if the slope is negative they are perpendicular.
2x - 3y = 1
2x + 3y = 2
Take x to the right
-3y = -2x + 1
3y = -2x + 2
Or
3y = 2x - 1
3y = -2x + 2
Divide both sides by 3
y = (2/3)x - 1
y = (-2/3)x - 1
Their slopes are negative. So they are perpendicular.
a. intersecting.
b. b= -3
c. x+(2/5)y=6/5
x-(a/3)y=4/3
Now two lines to be parallel,
-(a/3)=2/5
hence, a= -6/5
