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1. Distance, Slope of two points example
( Enter your problem )
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- Find the distance between the points A(5,-8) and B(-7,-3)
- Find the slope of the line joining points A(4,-8) and B(5,-2)
- If distance between the points (5,3) and (x,-1) is 5, then find the value of x
- If slope of the line joining points A(x,0), B(-3,-2) is 2/7, find the value of x
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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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3. If distance between the points (5,3) and (x,-1) is 5, then find the value of x
1. If distance between the points `A(5,3), B(x,-1)` is `5`, then find the value of `x`
Solution:
Points are `A(5,3),B(x,-1)` and distance`=5`
`:. x_1=5,y_1=3,x_2=x,y_2=-1` and `AB=5`
`AB=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
`=>5=sqrt((x-5)^2+(-1-3)^2)`
`=>25=x^2-10x+25+(16)`
`=>25=x^2-10x+41`
`=>x^2-10x+16 = 0`
`=>x^2-2x-8x+16 = 0`
`=>x(x-2)-8(x-2) = 0`
`=>(x-2)(x-8) = 0`
`=>(x-2) = 0" or "(x-8) = 0`
`=>x = 2" or "x = 8`
Hence value of x are 2 and 8
2. If distance between the points `A(x,-1), B(3,2)` is `5`, then find the value of `x`
Solution:
Points are `A(x,-1),B(3,2)` and distance`=5`
`:. x_1=x,y_1=-1,x_2=3,y_2=2` and `AB=5`
`AB=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
`=>5=sqrt((3-x)^2+(2--1)^2)`
`=>25=9-6x+x^2+(9)`
`=>25=x^2-6x+18`
`=>x^2-6x-7 = 0`
`=>x^2+x-7x-7 = 0`
`=>x(x+1)-7(x+1) = 0`
`=>(x+1)(x-7) = 0`
`=>(x+1) = 0" or "(x-7) = 0`
`=>x = -1" or "x = 7`
Hence value of x are -1 and 7
3. If distance between the points `A(x,2), B(3,-6)` is `10`, then find the value of `x`
Solution:
Points are `A(x,2),B(3,-6)` and distance`=10`
`:. x_1=x,y_1=2,x_2=3,y_2=-6` and `AB=10`
`AB=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
`=>10=sqrt((3-x)^2+(-6-2)^2)`
`=>100=9-6x+x^2+(64)`
`=>100=x^2-6x+73`
`=>x^2-6x-27 = 0`
`=>x^2+3x-9x-27 = 0`
`=>x(x+3)-9(x+3) = 0`
`=>(x+3)(x-9) = 0`
`=>(x+3) = 0" or "(x-9) = 0`
`=>x = -3" or "x = 9`
Hence value of x are -3 and 9
4. If distance between the points `A(x,1), B(-1,5)` is `5`, then find the value of `x`
Solution:
Points are `A(x,1),B(-1,5)` and distance`=5`
`:. x_1=x,y_1=1,x_2=-1,y_2=5` and `AB=5`
`AB=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
`=>5=sqrt((-1-x)^2+(5-1)^2)`
`=>25=1--2x+x^2+(16)`
`=>25=x^2--2x+17`
`=>x^2-2x-8 = 0`
`=>x^2+2x-4x-8 = 0`
`=>x(x+2)-4(x+2) = 0`
`=>(x+2)(x-4) = 0`
`=>(x+2) = 0" or "(x-4) = 0`
`=>x = -2" or "x = 4`
Hence value of x are -2 and 4
5. If distance between the points `A(x,7), B(1,15)` is `10`, then find the value of `x`
Solution:
Points are `A(x,7),B(1,15)` and distance`=10`
`:. x_1=x,y_1=7,x_2=1,y_2=15` and `AB=10`
`AB=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
`=>10=sqrt((1-x)^2+(15-7)^2)`
`=>100=1-2x+x^2+(64)`
`=>100=x^2-2x+65`
`=>x^2-2x-35 = 0`
`=>x^2+5x-7x-35 = 0`
`=>x(x+5)-7(x+5) = 0`
`=>(x+5)(x-7) = 0`
`=>(x+5) = 0" or "(x-7) = 0`
`=>x = -5" or "x = 7`
Hence value of x are -5 and 7
6. If distance between the points `A(1,x), B(-3,5)` is `5`, then find the value of `x`
Solution:
Points are `A(1,x),B(-3,5)` and distance`=5`
`:. x_1=1,y_1=x,x_2=-3,y_1=5` and `AB=5`
`AB=sqrt((x_2-x_1)^2+(y_2 - y_1)^2)`
`=>5=sqrt((-3 - 1)^2+(x-5)^2)`
`=>25=16+x^2-10x+25`
`=>25=x^2-10x+41`
`=>x^2-10x+16 = 0`
`=>x^2-2x-8x+16 = 0`
`=>x(x-2)-8(x-2) = 0`
`=>(x-2)(x-8) = 0`
`=>(x-2) = 0" or "(x-8) = 0`
`=>x = 2" or "x = 8`
Hence value of x are 2 and 8
7. If distance between the points `A(x,0), B(4,8)` is `10`, then find the value of `x`
Solution:
Points are `A(x,0),B(4,8)` and distance`=10`
`:. x_1=x,y_1=0,x_2=4,y_2=8` and `AB=10`
`AB=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`
`=>10=sqrt((4-x)^2+(8-0)^2)`
`=>100=16-8x+x^2+(64)`
`=>100=x^2-8x+80`
`=>x^2-8x-20 = 0`
`=>x^2+2x-10x-20 = 0`
`=>x(x+2)-10(x+2) = 0`
`=>(x+2)(x-10) = 0`
`=>(x+2) = 0" or "(x-10) = 0`
`=>x = -2" or "x = 10`
Hence value of x are -2 and 10
This material is intended as a summary. Use your textbook for detail explanation.
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