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**Distance of a Point from a Line**

Consider line L and point P in a coordinate plane.

The distance from point P to line L is equal to the length of perpendicular PM drawn from point P to line L. Let this distance be D.

Let line L be represented by the general equation of a line AX plus BY plus C is equal to zero.

Let the given line intersect the X- and the Y-axis at points Q and R, respectively.

The coordinates of point Q are obtained by solving the equation of the line and the equation of the X-axis, i.e. y=0.

On solving, the coordinates of point Q are (-C/A, 0).

The coordinates of point R are obtained by solving the equation of the line and the equation of the Y-axis, i.e. x=0.

On solving, the coordinates of point R are (0, -C/B).

Join PQ and PR to form triangle PQR.

Area of âˆ† = Â½ Base x Height

Area of âˆ†PQR = Â½ QR x PM â€¦â€¦ (1)

â‡’ PM = (2 x Area of Î”PQR)/QR â€¦.. (2)

If A (*x*_{1}, *y*_{1}), B (*x*_{2}, *y*_{2}) and C (*x*_{3}, *y*_{3}) form âˆ†ABC:

Area of âˆ†ABC = Â½ âˆ£*x*_{1}(*y*_{2} - *y*_{3}) + *x*_{2}(*y*_{3} - *y*_{1}) + *x*_{3}(*y*_{1} - *y*_{2})âˆ£

Given, P (*x*_{1}, *y*_{1}), Q (-C/A, 0) and R(0, -C/B)

Area of âˆ†PQR = Â½ âˆ£*x*_{1}{0 - (-C/B)} + (-C/A){(-C/B) - *y*_{1}} + 0{*y*_{1} - 0}âˆ£

â‡’ Area of âˆ†PQR = Â½ âˆ£*x*_{1}(C/B) + (-C/A) (-C/B) + *y*_{1}C/A + 0âˆ£

= Â½ âˆ£*x*_{1}C/B + C^{2}/AB + *y*_{1}C/Aâˆ£

= Â½ âˆ£C/ABâˆ£ x âˆ£A*x*_{1} + B*y*_{1} + Câˆ£ â€¦â€¦(3)

âˆ´ 2 x Area of âˆ†PQR = âˆ£C/ABâˆ£ x âˆ£A*x*_{1} + B*y*_{1} + Câˆ£ â€¦â€¦(4)

Given, points A (*x*_{1}, *y*_{1}) and B (*x*_{2}, *y*_{2})

Distance AB = âˆš(*x*_{2} - *x*_{1})^{2} + (*y*_{2} - *y*_{1})^{2}

Given, Q (-C/A, 0) and R (0, -C/B)

QR = âˆš{ (-C/A-0)}^{2} + {0-(-C/B) }^{2}

= âˆšC^{2}/A^{2} + C^{2}/B^{2}

= âˆš{(C^{2}/A^{2} B^{2} )(B^{2} + A^{2})

â‡’ QR = âˆ£C/ABâˆ£ âˆšA^{2} + B^{2} â€¦â€¦ (5)

â‡’ PM = [âˆ£C/ABâˆ£ x âˆ£A*x*_{1} + B*y*_{1} + Câˆ£]/[âˆ£C/ABâˆ£ âˆš[A^{2} + B^{2}]

â‡’ PM = âˆ£A*x*_{1} + B*y*_{1} + Câˆ£/âˆš[A^{2} + B^{2}]

â‡’ Distance (*d*) of a point P (*x*_{1}, *y*_{1}) from a line A*x* + B*y* + C = 0 is:

*d* = âˆ£A*x*_{1} + B*y*_{1} + Câˆ£/âˆš[A^{2} + B^{2}]

Distance between Two Parallel Lines

The distance between two parallel lines L_{1} and L_{2} in a coordinate plane:

If l_{1} âƒ¦ l_{2}: â‡’ Slope of l_{1} = Slope of l_{2} = m.

Represent line L_{1} as y = mx + c_{1}, and line L_{2} as y = mx + c_{2} in the slope-intercept form, respectively.

Let line L_{1} intersect the X-axis at point P.

Since point P lies on the X-axis, its y coordinate is equal to zero.

Substituting this value of y in the equation of line L_{1}, the x-coordinate of point P is -c_{1}/*m*.

Therefore, the coordinates of P are (-c_{1}/*m*, 0).

The distance between lines L1 and L2 is equal to the length of the perpendicular drawn from point P to line L2. Let this distance be 'd'.

Distance (*d*) of a point P (*x*_{1}, *y*_{1}) from a line A*x* + B*y* + C = 0 is

*d* = âˆ£A*x*_{1} + B*y*_{1} + Câˆ£/âˆšA^{2} + B^{2}]

*y* = *mx* + c_{2} â‡’ -*mx* + *y* - c_{2} = 0

Comparing -*mx* + *y* - c_{2} = 0 and A*x* + B*y* + C = 0

*A* = -*m*, B = 1 and C = -c_{2}

*â‡’* Distance *d* between point P (-c_{1}/*m*, 0) and line L2 is

*d = |(-m)(-c _{1}/m)+ 1 x 0 + (-c_{2})|/âˆš((-m)^{2} + 1^{2})*

â‡’ *d* = âˆ£c_{1} - c_{2}âˆ£/âˆš(1 + *m*^{2}) â€¦â€¦(1)

*â‡’* Distance between parallel lines *y* = *mx* + c_{1} and *y* = *mx* + c_{2} is

*d* = âˆ£c_{1} - c_{2}âˆ£/âˆš(1 + *m*^{2})

Suppose the equation of the parallel lines L1 and L2 are given in general form.

A*x* + B*y* + C_{1} = 0 â‡’ *y* = (-A/B)*x* +(- C_{1}/B) ...(2)

A*x* + B*y* + C_{2} = 0

â‡’ *y* = (-A/B)*x* +(- C_{2}/B) â€¦..(3)

Comparing equations 2 and 3 with *y* = *mx* + c, we get

â‡’ Slope (*m*) of lines L1 and L2 = -A/B

Also, c_{1} = -C_{1}/B and c_{2} = -C_{2}/B

Distance between parallel lines *y* = *mx* + c_{1} and *y* = *mx* + c_{2} is

*d* = âˆ£c_{1} - c_{2}âˆ£/âˆš(1 + *m*^{2})

â‡’ *d* = âˆ£(-C_{1}/B ) - (-C_{2}/B )âˆ£/âˆš(1 + (-A/B)^{2})

â‡’ *d* = âˆ£(C_{2}-C_{1})/B)âˆ£/âˆš(1 + A^{2}/B^{2})

= âˆ£(C_{2}-C_{1})/B âˆ£/âˆš(B^{2} + A^{2})/B^{2})

= âˆ£(C_{2}-C_{1})/B âˆ£/{âˆš(B^{2} + A^{2})}B

â‡’ *d* = âˆ£(C_{1}-C_{2})âˆ£/âˆš A^{2} + B^{2 }â€¦â€¦(4)

*â‡’* Distance between parallel lines A*x* + B*y* + C_{1} and A*x* + B*y* + C_{2} is

*d* = âˆ£(C_{1}-C_{2})âˆ£/âˆš A^{2} + B^{2}.

**Distance of a Point from a Line**

Consider line L and point P in a coordinate plane.

The distance from point P to line L is equal to the length of perpendicular PM drawn from point P to line L. Let this distance be D.

Let line L be represented by the general equation of a line AX plus BY plus C is equal to zero.

Let the given line intersect the X- and the Y-axis at points Q and R, respectively.

The coordinates of point Q are obtained by solving the equation of the line and the equation of the X-axis, i.e. y=0.

On solving, the coordinates of point Q are (-C/A, 0).

The coordinates of point R are obtained by solving the equation of the line and the equation of the Y-axis, i.e. x=0.

On solving, the coordinates of point R are (0, -C/B).

Join PQ and PR to form triangle PQR.

Area of âˆ† = Â½ Base x Height

Area of âˆ†PQR = Â½ QR x PM â€¦â€¦ (1)

â‡’ PM = (2 x Area of Î”PQR)/QR â€¦.. (2)

If A (*x*_{1}, *y*_{1}), B (*x*_{2}, *y*_{2}) and C (*x*_{3}, *y*_{3}) form âˆ†ABC:

Area of âˆ†ABC = Â½ âˆ£*x*_{1}(*y*_{2} - *y*_{3}) + *x*_{2}(*y*_{3} - *y*_{1}) + *x*_{3}(*y*_{1} - *y*_{2})âˆ£

Given, P (*x*_{1}, *y*_{1}), Q (-C/A, 0) and R(0, -C/B)

Area of âˆ†PQR = Â½ âˆ£*x*_{1}{0 - (-C/B)} + (-C/A){(-C/B) - *y*_{1}} + 0{*y*_{1} - 0}âˆ£

â‡’ Area of âˆ†PQR = Â½ âˆ£*x*_{1}(C/B) + (-C/A) (-C/B) + *y*_{1}C/A + 0âˆ£

= Â½ âˆ£*x*_{1}C/B + C^{2}/AB + *y*_{1}C/Aâˆ£

= Â½ âˆ£C/ABâˆ£ x âˆ£A*x*_{1} + B*y*_{1} + Câˆ£ â€¦â€¦(3)

âˆ´ 2 x Area of âˆ†PQR = âˆ£C/ABâˆ£ x âˆ£A*x*_{1} + B*y*_{1} + Câˆ£ â€¦â€¦(4)

Given, points A (*x*_{1}, *y*_{1}) and B (*x*_{2}, *y*_{2})

Distance AB = âˆš(*x*_{2} - *x*_{1})^{2} + (*y*_{2} - *y*_{1})^{2}

Given, Q (-C/A, 0) and R (0, -C/B)

QR = âˆš{ (-C/A-0)}^{2} + {0-(-C/B) }^{2}

= âˆšC^{2}/A^{2} + C^{2}/B^{2}

= âˆš{(C^{2}/A^{2} B^{2} )(B^{2} + A^{2})

â‡’ QR = âˆ£C/ABâˆ£ âˆšA^{2} + B^{2} â€¦â€¦ (5)

â‡’ PM = [âˆ£C/ABâˆ£ x âˆ£A*x*_{1} + B*y*_{1} + Câˆ£]/[âˆ£C/ABâˆ£ âˆš[A^{2} + B^{2}]

â‡’ PM = âˆ£A*x*_{1} + B*y*_{1} + Câˆ£/âˆš[A^{2} + B^{2}]

â‡’ Distance (*d*) of a point P (*x*_{1}, *y*_{1}) from a line A*x* + B*y* + C = 0 is:

*d* = âˆ£A*x*_{1} + B*y*_{1} + Câˆ£/âˆš[A^{2} + B^{2}]

Distance between Two Parallel Lines

The distance between two parallel lines L_{1} and L_{2} in a coordinate plane:

If l_{1} âƒ¦ l_{2}: â‡’ Slope of l_{1} = Slope of l_{2} = m.

Represent line L_{1} as y = mx + c_{1}, and line L_{2} as y = mx + c_{2} in the slope-intercept form, respectively.

Let line L_{1} intersect the X-axis at point P.

Since point P lies on the X-axis, its y coordinate is equal to zero.

Substituting this value of y in the equation of line L_{1}, the x-coordinate of point P is -c_{1}/*m*.

Therefore, the coordinates of P are (-c_{1}/*m*, 0).

The distance between lines L1 and L2 is equal to the length of the perpendicular drawn from point P to line L2. Let this distance be 'd'.

Distance (*d*) of a point P (*x*_{1}, *y*_{1}) from a line A*x* + B*y* + C = 0 is

*d* = âˆ£A*x*_{1} + B*y*_{1} + Câˆ£/âˆšA^{2} + B^{2}]

*y* = *mx* + c_{2} â‡’ -*mx* + *y* - c_{2} = 0

Comparing -*mx* + *y* - c_{2} = 0 and A*x* + B*y* + C = 0

*A* = -*m*, B = 1 and C = -c_{2}

*â‡’* Distance *d* between point P (-c_{1}/*m*, 0) and line L2 is

*d = |(-m)(-c _{1}/m)+ 1 x 0 + (-c_{2})|/âˆš((-m)^{2} + 1^{2})*

â‡’ *d* = âˆ£c_{1} - c_{2}âˆ£/âˆš(1 + *m*^{2}) â€¦â€¦(1)

*â‡’* Distance between parallel lines *y* = *mx* + c_{1} and *y* = *mx* + c_{2} is

*d* = âˆ£c_{1} - c_{2}âˆ£/âˆš(1 + *m*^{2})

Suppose the equation of the parallel lines L1 and L2 are given in general form.

A*x* + B*y* + C_{1} = 0 â‡’ *y* = (-A/B)*x* +(- C_{1}/B) ...(2)

A*x* + B*y* + C_{2} = 0

â‡’ *y* = (-A/B)*x* +(- C_{2}/B) â€¦..(3)

Comparing equations 2 and 3 with *y* = *mx* + c, we get

â‡’ Slope (*m*) of lines L1 and L2 = -A/B

Also, c_{1} = -C_{1}/B and c_{2} = -C_{2}/B

Distance between parallel lines *y* = *mx* + c_{1} and *y* = *mx* + c_{2} is

*d* = âˆ£c_{1} - c_{2}âˆ£/âˆš(1 + *m*^{2})

â‡’ *d* = âˆ£(-C_{1}/B ) - (-C_{2}/B )âˆ£/âˆš(1 + (-A/B)^{2})

â‡’ *d* = âˆ£(C_{2}-C_{1})/B)âˆ£/âˆš(1 + A^{2}/B^{2})

= âˆ£(C_{2}-C_{1})/B âˆ£/âˆš(B^{2} + A^{2})/B^{2})

= âˆ£(C_{2}-C_{1})/B âˆ£/{âˆš(B^{2} + A^{2})}B

â‡’ *d* = âˆ£(C_{1}-C_{2})âˆ£/âˆš A^{2} + B^{2 }â€¦â€¦(4)

*â‡’* Distance between parallel lines A*x* + B*y* + C_{1} and A*x* + B*y* + C_{2} is

*d* = âˆ£(C_{1}-C_{2})âˆ£/âˆš A^{2} + B^{2}.