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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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1. Find the equation of the line passing through the point A(5,4) and parallel to the line 2x+3y+7=0
1. Find the equation of the line passing through the point `A(5,4)` and parallel to the line `2x+3y+7=0`
Solution:
Here Point `(x_1,y_1)=(5,4)` and line `2x+3y+7=0` (given)
When two lines are parallel, their slopes are equal.
We shall first find the slope of line `2x+3y+7=0`
`2x+3y+7=0`
`:. 3y=-2x-7`
`:. y=-(2x)/(3)-7/3`
`:.` Slope `=-2/3` and hence slope of the line parallel to this line is also `-2/3`
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)=(5,4)` and Slope `m=-2/3` (given)
`:. y-4=-2/3(x-5)`
`:. 3(y-4)=-2(x-5)`
`:. 3y -12=-2x +10`
`:. 2x+3y-22=0`
Hence, The equation of line is `2x+3y-22=0`
2. Find the equation of the line passing through the point `A(1,1)` and parallel 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 parallel, their slopes are equal.
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` and hence slope of the line parallel to this line is also `2/3`
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=2/3` (given)
`:. y-1=2/3(x-1)`
`:. 3(y-1)=2(x-1)`
`:. 3y -3=2x -2`
`:. 2x-3y+1=0`
Hence, The equation of line is `2x-3y+1=0`
3. Find the equation of the line passing through the point `A(2,3)` and parallel to the line `2x-3y+8=0`
Solution:
Here Point `(x_1,y_1)=(2,3)` and line `2x-3y+8=0` (given)
When two lines are parallel, their slopes are equal.
We shall first find the slope of line `2x-3y+8=0`
`2x-3y+8=0`
`:. 3y=2x+8`
`:. y=(2x)/(3)+8/3`
`:.` Slope `=2/3` and hence slope of the line parallel to this line is also `2/3`
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)=(2,3)` and Slope `m=2/3` (given)
`:. y-3=2/3(x-2)`
`:. 3(y-3)=2(x-2)`
`:. 3y -9=2x -4`
`:. 2x-3y+5=0`
Hence, The equation of line is `2x-3y+5=0`
4. Find the equation of the line passing through the point `A(2,-5)` and parallel to the line `2x-3y-7=0`
Solution:
Here Point `(x_1,y_1)=(2,-5)` and line `2x-3y-7=0` (given)
When two lines are parallel, their slopes are equal.
We shall first find the slope of line `2x-3y-7=0`
`2x-3y-7=0`
`:. 3y=2x-7`
`:. y=(2x)/(3)-7/3`
`:.` Slope `=2/3` and hence slope of the line parallel to this line is also `2/3`
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)=(2,-5)` and Slope `m=2/3` (given)
`:. y+5=2/3(x-2)`
`:. 3(y+5)=2(x-2)`
`:. 3y +15=2x -4`
`:. 2x-3y-19=0`
Hence, The equation of line is `2x-3y-19=0`
This material is intended as a summary. Use your textbook for detail explanation.
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