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9. Find the equation of a line passing through a point and parallel or perpendicular to Line-2 or point-2 and point-3 example
( Enter your problem )
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- Find the equation of the line passing through the point A(5,4) and parallel to the line 2x+3y+7=0
- Find the equation of the line passing through the point A(1,1) and perpendicular to the line 2x-3y+2=0
- Find the equation of the line passing through the point A(1,3) and parallel to line passing through the points B(3,-5) and C(-6,1)
- Find the equation of the line passing through the point A(5,5) and perpendicular to the line passing through the points B(1,-2) and C(-5,2)
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Other related methods
- Distance, Slope of two points
- Points are Collinear or Triangle or Quadrilateral form
- Find Ratio of line joining AB and is divided by P
- Find Midpoint or Trisection points or equidistant points on X-Y axis
- Find Centroid, Circumcenter, Area of a triangle
- Find the equation of a line using slope, point, X-intercept, Y-intercept
- Find Slope, X-intercept, Y-intercept of a line
- Find the equation of a line passing through point of intersection of two lines and slope or a point
- Find the equation of a line passing through a point and parallel or perpendicular to Line-2 or point-2 and point-3
- Find the equation of a line passing through point of intersection of Line-1, Line-2 and parallel or perpendicular to Line-3
- For two lines, find Angle, intersection point and determine if parallel or perpendicular lines
- Reflection of points about x-axis, y-axis, origin
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2. Find the equation of the line passing through the point A(1,1) and perpendicular to the line 2x-3y+2=0
1. Find the equation of the line passing through the point `A(1,1)` and perpendicular to the line `2x-3y+2=0`
Solution:
Here Point `(x_1,y_1)=(1,1)` and line `2x-3y+2=0` (given)
When two lines are perpendicular, their slopes are opposite reciprocals of one another or the product of their slopes is -1.
We shall first find the slope of line `2x-3y+2=0`
`2x-3y+2=0`
`:. 3y=2x+2`
`:. y=(2x)/(3)+2/3`
`:.` Slope `=2/3`
`:.` Slope of perpendicular line`=(-1)/(2/3)=-3/2` (`:' m_1*m_2=-1`)
The equation of a line with slope m and passing through `(x_1,y_1)` is `y-y_1=m(x-x_1)`
Here Point `(x_1,y_1)=(1,1)` and Slope `m=-3/2` (given)
`:. y-1=-3/2(x-1)`
`:. 2(y-1)=-3(x-1)`
`:. 2y -2=-3x +3`
`:. 3x+2y-5=0`
Hence, The equation of line is `3x+2y-5=0`
2. Find the equation of the line passing through the point `A(0,0)` and perpendicular to the line `7x-3y+4=0`
Solution:
Here Point `(x_1,y_1)=(0,0)` and line `7x-3y+4=0` (given)
When two lines are perpendicular, their slopes are opposite reciprocals of one another or the product of their slopes is -1.
We shall first find the slope of line `7x-3y+4=0`
`7x-3y+4=0`
`:. 3y=7x+4`
`:. y=(7x)/(3)+4/3`
`:.` Slope `=7/3`
`:.` Slope of perpendicular line`=(-1)/(7/3)=-3/7` (`:' m_1*m_2=-1`)
The equation of a line with slope m and passing through `(x_1,y_1)` is `y-y_1=m(x-x_1)`
Here Point `(x_1,y_1)=(0,0)` and Slope `m=-3/7` (given)
`:. y-0=-3/7(x-0)`
`:. 7(y)=-3(x)`
`:. 7y +0=-3x +0`
`:. 3x+7y=0`
Hence, The equation of line is `3x+7y=0`
3. Find the equation of the line passing through the point `A(3,2)` and perpendicular to the line `y=5x+2`
Solution:
Here Point `(x_1,y_1)=(3,2)` and line `y=5x+2` (given)
When two lines are perpendicular, their slopes are opposite reciprocals of one another or the product of their slopes is -1.
We shall first find the slope of line `y=5x+2`
`y=5x+2`
`:. y=5x+2`
`:.` Slope `=5`
`:.` Slope of perpendicular line`=(-1)/(5)=-1/5` (`:' m_1*m_2=-1`)
The equation of a line with slope m and passing through `(x_1,y_1)` is `y-y_1=m(x-x_1)`
Here Point `(x_1,y_1)=(3,2)` and Slope `m=-1/5` (given)
`:. y-2=-1/5(x-3)`
`:. 5(y-2)=-1(x-3)`
`:. 5y -10=-x +3`
`:. x+5y-13=0`
Hence, The equation of line is `x+5y-13=0`
4. Find the equation of the line passing through the point `A(-1,1)` and perpendicular to the line `2x+3y+4=0`
Solution:
Here Point `(x_1,y_1)=(-1,1)` and line `2x+3y+4=0` (given)
When two lines are perpendicular, their slopes are opposite reciprocals of one another or the product of their slopes is -1.
We shall first find the slope of line `2x+3y+4=0`
`2x+3y+4=0`
`:. 3y=-2x-4`
`:. y=-(2x)/(3)-4/3`
`:.` Slope `=-2/3`
`:.` Slope of perpendicular line`=(-1)/(-2/3)=3/2` (`:' m_1*m_2=-1`)
The equation of a line with slope m and passing through `(x_1,y_1)` is `y-y_1=m(x-x_1)`
Here Point `(x_1,y_1)=(-1,1)` and Slope `m=3/2` (given)
`:. y-1=3/2(x+1)`
`:. 2(y-1)=3(x+1)`
`:. 2y -2=3x +3`
`:. 3x-2y+5=0`
Hence, The equation of line is `3x-2y+5=0`
This material is intended as a summary. Use your textbook for detail explanation.
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