Whats the equation of a line conainig point (-3,5) and parallel to line 4x-y=7?
2007-05-31 04:07:17 · 4 answers · asked by wltz4vns 1 in Science & Mathematics ➔ Mathematics
First put the line in slope intercept form: y=4x-7 from which you can see that the slope of this line is 4--therefore, the slope of the line parallel to this going through the point (-3,5) must also be 4. Using point slope I get: y-5=4(x+3) or y=4x+17.
2007-05-31 04:12:48 · answer #1 · answered by bruinfan 7 · 0⤊ 0⤋
Hello
Lets rearrange the equation into slope-intercept form.Thus we have y = 4x-7.
So the slope = 4. So we want the equation of the line parallel --- thus we will still use the slope 4.
Lets plug in the pt -3,5 into y = 4x+b to get the y-intercept.
5 = -12 + b ------ 17 = b.
Thus the equation is y = 4x+17.
Hope this helps
2007-05-31 11:14:02 · answer #2 · answered by Jeff U 4 · 0⤊ 0⤋
parallel means the same slope so m=4
using the point y= mx+b= 5 =4(-3) +b
b =17
so y= 4x+17
2007-05-31 11:14:16 · answer #3 · answered by xandyone 5 · 0⤊ 0⤋
m=4 for 4x-y=7
using m= y-y1/x-x1 substitute -3 for x1 5 for y1 therefore the equation becomes:
4= y-5/x-(-3) simplifying equation becomes 4x-y=-17
2007-05-31 11:19:24 · answer #4 · answered by Riza m 3 · 0⤊ 0⤋
m = y1 - y2/x1 -x2
2007-05-31 11:11:02 · answer #5 · answered by clawedstar 1 · 0⤊ 0⤋