The equations of two straight lines are = 3 + 13 and = 7 − 3.
Use algebra to solve these two simultaneous equations to find the co-ordinates of the point
where the lines meet.
Answers
Answer:
This line has some slope value (though not a value of "2", of course, because this line equation isn't solved for "y=").
I'll first need to find the slope of the reference line. I could use the method of twice plugging x-values into the reference line, finding the corresponding y-values, and then plugging the two points I'd found into the slope formula, but I'd rather just solve for "y=". (This is just my personal preference. If your preference differs, then use whatever method you like best.) So:
The first thing I'll do is solve "2x – 3y = 9" for "y=", so that I can find my reference slope:
2x - 3y = 92x−3y=9
3y = -2x + 93y=−2x+9
y = \dfrac{2}{3}x - 3y=
3
2
x−3
So the reference slope from the reference line is m = \frac{2}{3}m=
3
2
.
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Now I need to find two new slopes, and use them with the point they've given me; namely, with the point (4, –1). They want me to find the line through (4, –1) that is parallel to 2x – 3y = 9; that is, through the given point, they want me to find a line that has the same slope as the reference line. And they then want me to find the line through (4, –1) that is perpendicular to 2x – 3y = 9; that is, through the given point, they want me to find the line that has a slope which is the negative reciprocal of the slope of the reference line.
Since a parallel line has an identical slope, then the parallel line through (4, –1) will have slope m = \frac{2}{3}m=
3
2
. Hey, now I have a point and a slope! So I'll use the point-slope form to find the line:
y - (-1) = \dfrac{2}{3}(x - 4)y−(−1)=
3
2
(x−4)
y + 1 = \dfrac{2}{3}x - \dfrac{8}{3}y+1=
3
2
x−
3
8
y = \dfrac{2}{3}x - \dfrac{8}{3} - \dfrac{3}{3}y=
3
2
x−
3
8
−
3
3
y = \dfrac{2}{3}x - \dfrac{11}{3}y=
3
2
x−
3
11
This is the parallel line that they'd asked for, and it's in the slope-intercept form that they'd specified.
For the perpendicular line, I have to find the perpendicular slope. The reference slope is m = \frac{2}{3}m=
3
2
. For the perpendicular slope, I'll flip the reference slope and change the sign. Then the perpendicular slope is m = -\frac{3}{2}m=−
2
3
.
Again, I have a point and a slope, so I can use the point-slope form to find my equation. Note that the only change, in what follows, from the calculations that I just did above (for the parallel line) is that the slope is different, now being the slope of the perpendicular line.
y - (-1) = -\dfrac{3}{2}(x - 4)y−(−1)=−
2
3
(x−4)
y + 1 = -\dfrac{3}{2}x + 6y+1=−
2
3
x+6
y = -\dfrac{3}{2}x + 5y=−
2
3
x+5
Then the full solution to this exercise