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Find the Distance of the Point (1, 2) from the Mid-point of the Line Segment Joining the Points (6, 8) and (2, 4). - Mathematics

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प्रश्न

Find the distance of the point (1, 2) from the mid-point of the line segment joining the points (6, 8) and (2, 4).

उत्तर

We have to find the distance of a point A (1, 2) from the mid-point of the line segment joining P (6, 8) and Q (2, 4).

In general to find the mid-point P(x,y) of any two points `A(x_1, y_1)` and `B(x_2, y_2)` we use section formula as

`P(x,y) = ((x_1 + x_2)/2, (y_1 + y_2)/2)`

Therefore mid-point B of line segment PQ can be written as,

`B(x,y) = ((6 + 2)/2, (4 + 8)/2)`

Now equate the individual terms to get,

x = 4

y = 6

So co-ordinates of B is (4, 6)

Therefore distance between A and B,

`AB = sqrt((4 - 1)^2 + (6 - 2)^2)`

`= sqrt(9 + 16)`

=5

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अध्याय 6: Co-Ordinate Geometry - Exercise 6.3 [पृष्ठ २९]

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आरडी शर्मा Mathematics [English] Class 10
अध्याय 6 Co-Ordinate Geometry
Exercise 6.3 | Q 32 | पृष्ठ २९

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Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).


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The line segment joining the points (3, -4) and (1, 2) is trisected at the points P and Q. If the coordinates of P and Q are (p, -2) and (5/3, q) respectively. Find the values of p and q.


The line segment joining A(4, 7) and B(−6, −2) is intercepted by the y – axis at the point K. write down the abscissa of the point K. hence, find the ratio in which K divides AB. Also, find the co-ordinates of the point K.


Find the ratio in which the y-axis divides the line segment joining the points (−4, − 6) and (10, 12). Also find the coordinates of the point of division ?


Find the ratio in which the line segment joining A (2, -3) and B(S, 6) i~ divided by the x-axis. 


The points A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of ∆ABC. Find the coordinates of the point P on AD such that AP : PD = 2 : 1


Complete the following activity to find the coordinates of point P which divides seg AB in the ratio 3:1 where A(4, – 3) and B(8, 5).

Activity:

∴ By section formula,

∴ x = `("m"x_2 + "n"x_1)/square`, 

∴ x = `(3 xx 8 + 1 xx 4)/(3 + 1)`,

= `(square + 4)/4`,

∴ x = `square`,

∴ y = `square/("m" + "n")`

∴ y = `(3 xx 5 + 1 xx (-3))/(3 + 1)`

= `(square - 3)/4`

∴ y = `square`


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