If three lines whose equations are y\(_1\) = m\(_1\)x + c\(_1\), y = m\(_2\)x + c\(_2\) and y = m\(_3\)x + c\(_3\) are concurrent, then show that m\(_1\) (c\(_2\) - c\(_3\)) + m\(_2\) (c\(_3\) - c\(_1\)) + m\(_3\) (c\(_1\) - c\(_2\)) = 0

Solution:

The equations of the given lines are:

y = m\(_1\)x + c\(_1\) ....(1)

y = m\(_2\)x + c\(_2\) ....(2)

y = m\(_3\)x + c\(_3\) ....(3)

On subtracting equation (1) from (2) we obtain

(m\(_2\) - m\(_1\)) x + (c\(_2\) - c\(_1\)) = 0

(m\(_1\) - m\(_2\)) x = (c\(_2\) - c\(_1\))

x = (c\(_2\) - c\(_1\))/(m\(_1\) - m\(_2\))

On substituting this value of x in (1), we obtain

y = m\(_1\) [(c\(_2\) - c\(_1\))/(m\(_1\) - m\(_2\))] + c\(_1\)

= (m\(_1\)c\(_2\) - m\(_1\)c\(_1\))/(m\(_1\) - m\(_2\))] + c\(_1\)

= (m\(_1\)c\(_2\) - m\(_1\)c\(_1\) + m\(_1\)c\(_1\) - m\(_2\)c\(_1\))/(m\(_1\) - m\(_2\))

= (m\(_1\)c\(_2\) - m\(_2\)c\(_1\))/(m\(_1\) - m\(_2\))

Therefore,

[(c\(_2\) - c\(_1\))/(m\(_1\) - m\(_2\)), (m\(_1\)c\(_2\) - m\(_2\)c\(_1\))/(m\(_1\) - m\(_2\))] is the point of intersection of line (1) and (2)

It is given that lines (1), (2) and (3) are concurrent.

Hence the point of intersection of lines (1) and (2) will also satisfy equation (3).

On substituting this value of x and y in (3), we obtain

(m\(_1\)c\(_2\) - m\(_2\)c\(_1\))/(m\(_1\) - m\(_2\)) = m\(_3\)(c\(_2\) - c\(_1\))/(m\(_1\) - m\(_2\)) + c\(_3\)

(m\(_1\)c\(_2\) - m\(_2\)c\(_1\))/(m\(_1\) - m\(_2\)) = (m\(_3\)c\(_2\) - m\(_3\)c\(_1\) + c\(_3\)m\(_1\) - c\(_3\)m\(_2\))/(m\(_1\) - m\(_2\))

m\(_1\)c\(_2\) - m\(_2\)c\(_1\) - m\(_3\)c\(_2\) + m\(_3\)c\(_1\) - c\(_3\)m\(_1\) + c\(_3\)m\(_2\) = 0

m\(_1\) (c\(_2\) - c\(_3\)) + m\(_2\) (c\(_3\) - c\(_1\)) + m\(_3\) (c\(_1\) - c\(_2\)) = 0

Hence, m\(_1\) (c\(_2\) - c\(_3\)) + m\(_2\) (c\(_3\) - c\(_1\)) + m\(_3\) (c\(_1\) - c\(_2\)) = 0 is shown

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NCERT Solutions Class 11 Maths Chapter 10 Exercise ME Question 10

If three lines whose equations are y\(_1\) = m\(_1\)x + c\(_1\), y = m\(_2\)x + c\(_2\) and y = m\(_3\)x + c\(_3\) are concurrent, then show that m\(_1\) (c\(_2\) - c\(_3\)) + m\(_2\) (c\(_3\) - c\(_1\)) + m\(_3\) (c\(_1\) - c\(_2\)) = 0

Summary:

If three lines whose equations are y\(_1\) = m\(_1\)x + c\(_1\), y = m\(_2\)x + c\(_2\) and y = m\(_3\)x + c\(_3\) are concurrent, then we are asked to show that m\(_1\) (c\(_2\) - c\(_3\)) + m\(_2\) (c\(_3\) - c\(_1\)) + m\(_3\) (c\(_1\) - c\(_2\)) = 0 and we have shown it