inding the equation of a straight line y=mx+c which is perpendicular to the line 3x-4y+5=0 and passes through the point of intersection of the lines 2x+y=5 and x-y= just say hi for this point
Answers
Step-by-step explanation:
Step 1: Put the two equations in y=mx+b format since that is the formula of a straight line.
2x+3y-5= 0 -> y=-2/3x+5/3
3x+4y-7=0 -> y=-3/4x+7/4
Then to find the intersection of those two straight line equations, I would do the process of substitution or elimination method to find which coordinate pair( x,y) is being intersected by the two equation. ( Faster way would be just plugging the two equations in a graph and find the point where they intersect)
I think Subsitution would be easier because we’ve already solved for y bc of the format y=mx+b:
-2/3x+5/3=-3/4x+7/4
Solve for x:
Step 1: Get common denominators or LCM( least common multiple). That would be 12.
-8/12x+20/12= -9/12x+21/12
Step 2: Get x on one side of the equation. I added 8/12x.
20/12= -1/12x+21/12
Step 3: subtract 21/12
-1/12= -1/12x
Then Divide by -1/12 and x = 1
Now that x has been solved, substitute it into one of the equations above. I substituted x in the equation y=-3/4x+7/4
y=-3/4(1)+ 7/4
y=4/4
y= 1
So the intersection between the two equations is at the coordinate pair of (1,1)
When finding a line that is perpendicular to the other line. The product of their slopes must be equal to -1.
So 5x-3y-2=0 in y=mx+b form is y=5/3x -2/3
Just based on that information, we know that the slope of the equation we’re trying to make has to be -3/5 bc -3/5*5/3= -1.
Plug the coordinate pair (1,1) into this new equation:
1=-3/5(1)+ b
Solve for b.
b= 8/5
So the new equation that has a point at the intersection of (1,1) from the first two equations and is perpendicular to 5x-3y-2=0 is y=-3/5x+8/5.