
Anamta J. answered 06/15/19
Certified 6-12 Math Teacher and Middle School Math Interventionist
- We will use the point-slope form of a linear equation in this case,since we are given two points lying on the line.
y-y1=m(x-x1)
y-3=m(x-7)
need to calculate m where m is the gradient .
m=(y2-y1)/(x2-x1)
plug in the values for each coordinate. P1(x1,y1) P2 (x2,y2)
m=(-6 -3)/(2-7) = -9/5
y-3=-9/5 (x-7)
y=-9/5x+63/5 +3
= -9/5x+78/5
2.The slope and a point is already given so we just plug the values in the general point-slope equation.
y-y1=m(x-x1)
y-3=-1/3(x-2)
Solve the equation for y
y= -1/3x+2/3+3
y=-1/3x+11/3
3.The slope and a point is already given so we just plug the values in the general point-slope equation.x-intercept means a point ( x,0) so we have a point(-3/4,0)
y-y1=m(x-x1)
y-0=-4(x- (-3/4))
y=-4(x+3/4)
y=-4x-3
4 . We use the y=mx+c form of the linear equation since the intercept is given
y=mx+c where m is the gradient and c is the intercept
simply plug in the values for m and c and you're done
y=2/3x+4
we can also use the procedure used in part(3) by using (0,4) as our point.
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Find the equation of a line.
1. parallel to y = 3x + 4, passing through P1(-2, 5)
parallel lines have the same gradient, so we know our line y=mx+c or y-y1=m(x-x1) has m=3.
Using the point-slope form find the equation.
y-y1=m(x-x1)
y-5=3(x-(-2))
y-5=3(x+2)
y=3x+6+5
y=3x+11
2. parallel to a line passing P1(0, -3) and P2(-1, 4) and passing through P3(-1, 4).
the asked line has the same gradient as the line passing through P1 and P2 so we find the gradient
m=(y2-y1)/(x2-x1)
m=(4-(-3))/(-1) = -7
Now Using the point-slope form find the equation.
y-y1=m(x-x1)
y-4=-7(x-(-1))
y-4=-7(x+1)
y=-7x-7+4
y=-7x-3
3. perpendicular to the line y = x + 4 and passing through (-2, -6).
The gradient of the given line is 1.
The product of the gradient of two perpendicular lines is -1.
so the gradient of the asked line is -1.
Now using the point-slope form find the equation.
y-y1=m(x-x1)
y-(-6)=-1(x-(-2))
y+6=-1(x+2)
y=-x-2-6
y=-x-8
4. perpendicular to the line passing through point P1(3, 2) , P2(4, -6), and passing through P3(-6, 4).
Finding the gradient of the given line.
m=(y2-y1)/(x2-x1)
m=(-6-2)/(4-3) = -8
so the gradient of the required line is 1/8
Now using the point-slope form find the equation.
y-y1=m(x-x1)
y-(4)=1/8(x-(-6))
y-4=1/8(x+6)
y=1/8x + 6/8+4
y=1/8x +19/4
Find the distance between P1(3, 4) , P2(-2, -6).
distance between two points is given by formula √(x2-x1)2+(y2-y1)2
the entire formula is under the square root
√(-2-3)2+(-6-4)2 = √25+100 =√125 =11.18 cm