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Yogita Ingle 6 years, 4 months ago
we know that equation of line
y = mx + c
where c is constant
{tex}m = \frac{y2 - y1}{x2 - x1} \\ m = \frac{6 - ( - 3)}{5 - ( - 2)} \\ m = \frac{9}{7}{/tex}
y = (9/7)x + c
when x = 5 then y = 6
6 = (9/7)×5 + c
=> 42 = 45 + 7c
=> c = -3/7
equation is
y = (9/7)x - 3/7
or
7y = 9x - 3
x axis interception means y = 0
=> 0 = 9x -3
=> x = 3/9 = 1/3
y axis interception means x = 0
=> 7y = -3
=> y = -3/7
y axis intercepted at (0, -3/7)
Distance from (-2,-3)
{tex}= \sqrt{(-2-0)^{2} + (-3 -(-3/7)^2} = \sqrt{(-2)^2 + (\frac{-18}{7})^2 } \\= \sqrt{4 + \frac{324}{49} } = \sqrt{\frac{520}{49}}{/tex}
Distance from (5, 6)
{tex}= \sqrt{(5-0)^{2} + (6 -(-3/7)^2} = \sqrt{(5)^2 + (\frac{45}{7})^2 } \\= \sqrt{25 + \frac{2025}{49} } = \sqrt{\frac{3250}{49}}{/tex}
{tex}Ratio = \sqrt{\frac{3250}{49}} / \sqrt{\frac{520}{49}} = \sqrt{6.25} = 2.5{/tex}
x axis interception at (1/3 , 0)
{tex}Distance from (-2,-3) = \sqrt{(-2- \frac{1}{3})^2 + (-3-0)^2} = \sqrt{\frac{49}{9} + 9} = \sqrt{\frac{130}{9}}{/tex}
{tex}Distance from (5,6) = \sqrt{(5- \frac{1}{3})^2 + (6-0)^2} = \sqrt{\frac{196}{9} + 36} = \sqrt{\frac{520}{9}}{/tex}
{tex}Ratio = \sqrt{\frac{520}{130}} = \sqrt{4} = 2{/tex}
Posted by Himanshi Thakkar 6 years, 4 months ago
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Gaurav Seth 6 years, 4 months ago
Q : The sum of n, 2n ,3n terms of an AP are S1,S2,S3,respectively. Prove that S3 =3(S2- S1)
Answer:
Let ‘a’ be the first term of the AP and ‘d’ be the common difference
S1 = (n/2)[2a + (n – 1)d] --- (1)
S2 = (2n/2)[2a + (2n – 1)d] = n[2a + (n – 1)d] --- (2)
S3 = (3n/2)[2a + (3n – 1)d] --- (3)
Consider the RHS: 3(S2 – S1)
= S3
= L.H.S
∴ S3 = 3(S2 - S1) .
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Yogita Ingle 6 years, 4 months ago
A = (-1, 7) and B = (4, -3)
So, x1=−1, y1=7, x2=4 and y2=−3
We also have; m1=2 and m2=3
Let us assume P is the point of division
Coordinates of P can be calculated as follows by using section formula:
Hence, P = (1, 3)
Posted by Karthik A 6 years, 4 months ago
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Sahil ???? 6 years, 4 months ago
1Thank You