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Home > Geometry calculators > Coordinate Geometry > Find the value of x, If distance between the points (5,3) and (x,-1) is 5 calculator
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Method and examples
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Method
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Find the value of x, if the distance between the points (x,-1) and (3,2) is 5
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1. Distance, Slope of two points
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1. Find the distance between the points `A(5,-8)` and `B(-7,-3)`
2. Find the slope of the line joining points `A(4,-8)` and `B(5,-2)`
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` - `A(5,-8),B(-7,-3)`
- `A(7,-4),B(-5,1)`
- `A(-6,-4),B(9,-12)`
- `A(1,-3),B(4,-6)`
- `A(-5,7),B(-1,3)`
- `A(-8,6),B(2,0)`
- `A(0,0),B(7,4)`
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Find the value of x or y
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3. If distance between the points (5,3) and (x,-1) is 5, then find the value of x.
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Distance =
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- `A(5,3),B(x,-1)`, distance `=5`
- `A(x,-1),B(3,2)`, distance `=5`
- `A(x,2),B(3,-6)`, distance `=10`
- `A(x,1),B(-1,5)`, distance `=5`
- `A(x,7),B(1,15)`, distance `=10`
- `A(1,x),B(-3,5)`, distance `=5`
- `A(x,0),B(4,8)`, distance `=10`
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4. 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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Slope =
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- `A(x,0),B(-3,-2)`, slope `=2/7`
- `A(2,x),B(-3,7)`, slope `=1`
- `A(x,5),B(-1,2)`, slope `=3/4`
- `A(2,5),B(x,3)`, slope `=2`
- `A(x,2),B(6,-8)`, slope `=-5/4`
- `A(-2,x),B(5,-7)`, slope `=-1`
- `A(2,3),B(x,6)`, slope `=3/5`
- `A(-3,4),B(5,x)`, slope `=-5/4`
- `A(0,x),B(5,-2)`, slope `=-9/5`
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2. Points are Collinear or Triangle or Quadrilateral form
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Show that the points are the vertices of
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Find `A(0,0), B(2,2), C(0,4), D(-2,2)` are vertices of a square or not
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- `A(1,5),B(2,3),C(-2,-11)` are collinear points
- `A(1,-3),B(2,-5),C(-4,7)` are collinear points
- `A(-1,-1),B(1,5),C(2,8)` are collinear points
- `A(0,-1),B(3,5),C(5,9)` are collinear points
- `A(2,8),B(1,5),C(0,2)` are collinear points
- `A(-1,-1),B(1,5),C(2,8)` are collinear points
- `A(0,-1),B(3,5),C(5,9)` are collinear points
- `A(2,8),B(1,5),C(0,2)` are collinear points
- `A(0,0),B(0,3),C(4,0)` are vertices of a right angle triangle
- `A(-2,-2),B(-1,2),C(3,1)` are vertices of a right angle triangle
- `A(-3,2),B(1,2),C(-3,5)` are vertices of a right angle triangle
- `A(2,5),B(8,5),C(5,10.196152)` are vertices of an equilateral triangle
- `A(2,2),B(-2,4),C(2,6)` are vertices of an isosceles triangle
- `A(0,0),B(2,0),C(-4,0),D(-2,0)` are collinear points
- `A(3,2),B(5,4),C(3,6),D(1,4)` are vertices of a square
- `A(0,0),B(2,2),C(0,4),D(-2,2)` are vertices of a square
- `A(1,-1),B(-2,2),C(4,8),D(7,5)` are vertices of a rectangle
- `A(0,-4),B(6,2),C(3,5),D(-3,-1)` are vertices of a rectangle
- `A(3,0),B(4,5),C(-1,4),D(-2,-1)` are vertices of a rhombus
- `A(2,3),B(7,4),C(8,7),D(3,6)` are vertices of a parallelogram
- `A(1,5),B(1,4),C(-1,3),D(-1,4)` are vertices of a parallelogram
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3. Find Ratio of line joining AB and is divided by P
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1. Find the ratio in which the point P(3/4, 5/12) divides the line segment joining the points A(1/2, 3/2) and B(2, -5)
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- `P(3/4,5/12),A(1/2,3/2),B(2,-5)`
- `P(-1,6),A(3,10),B(6,-8)`
- `P(-2,3),A(-3,5),B(4,-9)`
- `P(3,10),A(5,12),B(2,9)`
- `P(6,17),A(1,-3),B(3,5)`
- `P(12,23),A(2,8),B(6,14)`
- `P(3,10),A(5,12),B(2,9)`
- `P(6,17),A(1,-3),B(3,5)`
- `P(12,23),A(2,8),B(6,14)`
- `P(17/5,47/5),A(5,13),B(1,4)`
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2. Write down the co-ordinates of the point P that divides the line joining A(-4,1) and B(17,10) in the ratio 1:2
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ratio =
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- `A(5,13),B(1,4),m:n=2:3`
- `A(-4,1),B(17,10),m:n=1:2`
- `A(5,12),B(2,9),m:n=2:1`
- `A(2,8),B(6,14),m:n=5:3` Externally
- `A(1,-3),B(3,5),m:n=5:3` Externally
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3. In what ratio does the x-axis divide the join of `A(2,-3)` and `B(5,6)`? Also find the coordinates of the point of intersection.
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divided by
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- `A(2,-3),B(5,6)` divided by x-axis
- `A(1,2),B(2,3)` divided by x-axis
- `A(5,-6),B(-1,-4)` divided by y-axis
- `A(-2,1),B(4,5)` divided by y-axis
- `A(2,1),B(7,6)` divided by x-axis
- `A(2,-4),B(-3,6)` divided by y-axis
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4. Find the ratio in which the point `P(x,4)` divides the line segment joining the points `A(2,1)` and `B(7,6)`? Also find the value of `x`.
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- `P(x,2),A(12,5),B(4,-3)`
- `P(11,y),A(15,5),B(9,20)`
- `P(-3,y),A(-5,-4),B(-2,3)`
- `P(-4,y),A(-6,10),B(3,-8)`
- `P(x,4),A(2,1),B(7,6)`
- `P(x,0),A(2,-4),B(-3,6)`
- `P(0,y),A(2,-4),B(-3,6)`
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4. Find Midpoint or Trisection points or equidistant points on X-Y axis
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1. Find the coordinates of the midpoint of the line segment joining the points `A(-5,4)` and `B(7,-8)`
2. Find the trisectional points of line joining `A(-3,-5)` and `B(-6,-8)`
3. Find the point on the x-axis which is equidistant from `A(5,4)` and `B(-2,3)`
4. Find the point on the y-axis which is equidistant from `A(6,5)` and `B(-4,3)`
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- `A(-5,4),B(7,-8)`
- `A(2,1),B(1,-3)`
- `A(2,1),B(5,3)`
- `A(3,-5),B(1,1)`
- `A(1,-1),B(-5,-3)`
- `A(-7,-3),B(5,3)`
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5. Find Centroid, Circumcenter, Area of a triangle
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1. Find the centroid of a triangle whose vertices are `A(4,-6),B(3,-2),C(5,2)`
2. Find the circumcentre of a triangle whose vertices are `A(-2,-3),B(-1,0),C(7,-6)`
3. Using determinants, find the area of the triangle with vertices are `A(-3,5),B(3,-6),C(7, 2)`
4. Using determinants show that the following points are collinear `A(2,3),B(-1,-2),C(5,8)`
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- `A(4,-6),B(3,-2),C(5,2)`
- `A(3,-5),B(-7,4),C(10,-2)`
- `A(4,-8),B(-9,7),C(8,13)`
- `A(3,-7),B(-8,6),C(5,10)`
- `A(2,4),B(6,4),C(2,0)`
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6. Find the equation of a line using slope, point, X-intercept, Y-intercept
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1. Find the equation of a straight line passing through `A(-4,5)` and having slope `-2/3`
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Slope :
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- `A(-4,5)`,slope`=-2/3`
- `A(4,5)`,slope`=1`
- `A(-2,3)`,slope`=-4`
- `A(-1,2)`,slope`=-5/4`
- `A(0,3)`,slope`=2`
- `A(0,0)`,slope`=1/4`
- `A(5,4)`,slope`=1/2`
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2. Find the equation of a straight line passing through the points `A(7,5)` and `B(-9,5)`
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- `A(7,5),B(-9,5)`
- `A(-1,1),B(2,-4)`
- `A(-5,-6),B(3,10)`
- `A(3,-5),B(4,-8)`
- `A(-1,-4),B(3,0)`
- `A(7,8),B(1,0)`
- `A(6,4),B(-1,5)`
- `A(2,3),B(7,6)`
- `A(-3,4),B(5,-6)`
- `A(0,7),B(5,-2)`
- `A(0,0),B(-4,-6)`
- `A(3,5),B(6,4)`
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4. Find the slope, x-intercept and y-intercept of the line joining the points `A(1,3)` and `B(3,5)`
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- `A(1,3),B(3,5)`
- `A(4,-8),B(5,-2)`
- `A(7,1),B(8,9)`
- `A(4,8),B(5,5)`
- `A(7,8),B(1,0)`
- `A(6,4),B(-1,5)`
- `A(2,3),B(7,6)`
- `A(-3,4),B(5,-6)`
- `A(0,7),B(5,-2)`
- `A(0,0),B(-4,-6)`
- `A(3,5),B(6,4)`
- `A(3,-5),B(-7,9)`
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8. Find the equation of a line passing through point of intersection of two lines and slope or a point
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1. Find the equation of a line passing through the point of intersection of lines `3x+4y=7` and `x-y+2=0` and having slope 5
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Line-1 : ,
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Line-2 : ,
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Slope :
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- Line-1`:x-4y+18=0`,Line-2`:x+y-12=0`,slope`=2`
- Line-1`:2x+3y+4=0`,Line-2`:3x+6y-8=0`,slope`=2`
- Line-1`:x=3y`,Line-2`:3x=2y+7`,slope`=-1/2`
- Line-1`:x-4y+18=0`,Line-2`:x+y-12=0`,slope`=2`
- Line-1`:2x+3y+4=0`,Line-2`:3x+6y-8=0`,slope`=2`
- Line-1`:x=3y`,Line-2`:3x=2y+7`,slope`=-1/2`
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2. Find the equation of a line passing through the point of intersection of lines `4x+5y+7=0` and `3x-2y-12=0` and point `A(3,1)`
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Line-1 : ,
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Line-2 : ,
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- Line-1`:x+y+1=0`,Line-2`:3x+y-5=0`,`A(1,-3)`
- Line-1`:4x+5y+7=0`,Line-2`:3x-2y-12=0`,`A(3,1)`
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9. Find the equation of a line passing through a point and parallel or perpendicular to Line-2
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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`
2. Find the equation of the line passing through the point `A(1,1)` and perpendicular to the line `2x-3y+2=0`
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Line-2 :
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- `A(5,4)`,Line`:2x+3y+7=0`
- `A(1,1)`,Line`:2x-3y+2=0`
- `A(2,3)`,Line`:2x-3y+8=0`
- `A(2,-5)`,Line`:2x-3y-7=0`
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3. 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)`
4. 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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- `A(1,3),B(3,-5),C(-6,1)`
- `A(4,-5),B(3,7),C(-2,4)`
- `A(-1,3),B(0,2),C(4,5)`
- `A(2,-3),B(1,2),C(-1,5)`
- `A(4,2),B(1,-1),C(3,2)`
- `A(5,5),B(1,-2),C(-5,2)`
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12. Reflection of points about x-axis, y-axis, origin
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Find Reflection of points A(0,0),B(2,2),C(0,4),D(-2,2) and Reflection about X,Y,O
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Reflection about
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- `A(-2,-2),B(-1,2),C(3,1)` and Reflection about x
- `A(2,3),B(7,4),C(8,7),D(3,6)` and Reflection about y
- `A(1,-1),B(-2,2),C(4,8),D(7,5)` and Reflection about o
- `A(3,0),B(4,5),C(-1,4),D(-2,-1)` and Reflection about x,y
- `A(3,2),B(5,4),C(3,6),D(1,4)` and Reflection about y,x
- `A(-1,-1),B(1,5),C(2,8)` and Reflection about y=x
- `A(-3,2),B(1,2),C(-3,5)` and Reflection about y=-x
- `A(0,-1),B(3,5),C(5,9)` and Reflection about x=2
- `A(2,8),B(1,5),C(0,2)` and Reflection about y=2
- `A(0,0),B(2,2),C(0,4),D(-2,2)` and Reflection about x+3y-7=0
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Solution
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Solution provided by AtoZmath.com
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Find the value of x, if the distance between the points (x,-1) and (3,2) is 5 calculator
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1. If distance between the points `A(5,3), B(x,-1)` is `5`, then find the value of `x`
2. If distance between the points `A(x,-1), B(3,2)` is `5`, then find the value of `x`
3. If distance between the points `A(x,2), B(3,-6)` is `10`, then find the value of `x`
4. If distance between the points `A(x,1), B(-1,5)` is `5`, then find the value of `x`
5. If distance between the points `A(x,7), B(1,15)` is `10`, then find the value of `x`
6. If distance between the points `A(1,x), B(-3,5)` is `5`, then find the value of `x`
7. If distance between the points `A(x,0), B(4,8)` is `10`, then find the value of `x`
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Example1. 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
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