anonymous
  • anonymous
A line passes through point (3, 7) and has a slope of 3/4
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
pooja195
  • pooja195
in what form?
anonymous
  • anonymous
y=mx+b slope y-ntercept

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anonymous
  • anonymous
Let the required line be y = mx+c in the slope intercept form, where m is the slope and c is the y intercept. We determine m and c by the using given conditions. m = 3/4, given. So y = (3/4)x+c (1). The line at (1) has the point (-8, 4) on it. Therefore 4 = (3/4)*(-8)+c...(2) (1)-(2) gives: y-4 = (3/4)(x+8). => y -4 = (3/4)x + 6 We multiply 4 to get integral coefficients. 4(y-4) = 3x+24 We rearrange. 3x-4y+24+16= 0 3x-4y+40 = 0 is the required line.
anonymous
  • anonymous
the options are- y=3/4x+19/4 and y=3x-19
anonymous
  • anonymous
what do you think it would be the answer?
anonymous
  • anonymous
y=3/4x+19/4
anonymous
  • anonymous
but im not sure
pooja195
  • pooja195
An easier approach for me atlest is this way: \[\huge y-y_1=m(x-x_1)\] where m=slope (x1,y1) (3, 7) and has a slope of 3/4 y-7=3/4(x-3) from here after you distribute and solve for y you get y=3/4x+19/4
pooja195
  • pooja195
So what you said is correct
anonymous
  • anonymous
sweet
anonymous
  • anonymous
If point A(x, 5) lies on the line, the value of x is 33 or 1/3
anonymous
  • anonymous
still need help @harryhart
anonymous
  • anonymous
anonymous
  • anonymous
open a new question

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