Solution:
Let the coordinates of the point be P(x, y) which divides the line segment joining the points (-1, 7) and (4, - 3) in the ratio 2 : 3
Let two points be A (x₁, y₁) and B(x₂, y₂). P (x, y) divides internally the line joining A and B in the ratio m₁: m₂. Then, coordinates of P(x, y) is given by the section formula
P (x, y) = [(mx₂ + nx₁ / m + n), (my₂ + ny₁ / m + n)]
Let x₁ = - 1, y₁ = 7, x₂ = 4 and y₂ = - 3, m = 2, n = 3
By Section formula, P (x, y) = [(mx₂ + nx₁ / m + n) , (my₂ + ny₁ / m + n)] --- (1)
By substituting the values in the equation (1)
x = [2 × 4 + 3 × (- 1)] / (2 + 3) and y = [2 × (- 3) + 3 × 7] / (2 + 3)
x = (8 - 3) / 5 and y = (- 6 + 21) / 5
x = 5/5 = 1 and y = 15/5 = 3
Therefore, the coordinates of point P are (1, 3).
☛ Check: NCERT Solutions Class 10 Maths Chapter 7
Video Solution:
Find the coordinates of the point which divides the join of (-1, 7) and (4, - 3) in the ratio 2 : 3
NCERT Class 10 Maths Solutions Chapter 7 Exercise 7.2 Question 1
Summary:
The coordinates of the point which divides the join of (- 1, 7) and (4, - 3) in the ratio 2 : 3 is (1, 3).
☛ Related Questions:
- Find the coordinates of the points of trisection of the line segment joining (4, - 1) and (- 2, - 3).
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Question 4 Section Formula Exercise 11
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Answer:
Solution:
In coordinate geometry, the Section formula is used to determine the internal or external ratio at which a line segment is divided by a point.
Let P(p,-2) and Q(5/3, q) be the points of trisection of AB
i.e., AP = PQ = QB
Given A(3,-4) and B(1,2)
x1 = 3, y1 = -4, x2 = 1, y2 = 2
P(p, -2) divides AB internally in the ratio 1 : 2.
By Section formula,
x=\frac{\left(mx_2+nx_1\right)}{(m+n)}\\ p=\frac{\left(1\times1+2\times3\right)}{(1+2)}\\ p=\frac{\left(1+6\right)}{3}\\ p=\frac{7}{3}
Now, Q also divides AB internally in the ratio 2 : 1.
m:n = 2:1
Q(5/3, q) divides AB internally in the ratio 2 : 1.
By Section formula,
y=\frac{\left(my2+ny1\right)}{(m+n)}\\ q=\frac{\left(2\times2+1\times-4\right)}{(2+1)}\\ q=\frac{\left(4-4\right)}{3}\\q=\frac{0}{3}\\ q=0
Hence the value of p and q are 7/3 and 0 respectively.
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Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.
Let P, Q and R be the three points which divide the line-segment joining the points A(-2, 2) and B(2, 8) in four equal parts.
Case I. For point P, we have
Hence, m1 = 1, m2 = 3
x1 = -2, y2 = 2
x2 = 2, y2 = 8Then, coordinates of P are given by
Case II. For point Q, we have
m1 = 2, m2 = 2
x1 = -2, y1 = 2
and x2 = 2, y2 = 8Then, coordinates of Q are given by
Case III. For point R, we have
Hence, m1 = 3, m2 = 1
x1 = -2, y1 = 2
and x2 = 2, y2 = 8Then co-ordinates of R are given by
Text Solution
Solution : Let `A(1, -2)` and `B(-3, 4)` be the given points. <br> Let the points of trisection be P and Q. <br> Then `AP = PQ = QB = lambda`(say)<br> `PB = PQ + QB = 2lambda`<br> and `AQ = AP + PQ = 2lamda`<br> `AP : PB =1:2` and `AQ : QB =2:1`<br> So P divides AB internally in the ratio `1:2` while Q divides internally in the ratio `2:1`<br> thus,the co-ordinates of P are <br>`(1xx−3+2xx1)/(1+2),(1xx4+2xx−2)/(1+2)`or`(−1/3,0)`<br> and the co-ordinates of Q are <br>`(2xx−3+1xx1)/(2+1),(2xx4+1xx(−2))/(2+1)`or`(−5/3,2)`<br> Hence, the points of trisection are `(−1/3,0)`and`(−5/3,2)`
Find the coordinates of points which trisect the line segment joining 1, 2 and 3,4.
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