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प्रश्न
Show that lines 3x − 4y + 5 = 0, 7x − 8y + 5 = 0, and 4x + 5y − 45 = 0 are concurrent. Find their point of concurrence
उत्तर
The number of lines intersecting at a point are called concurrent lines and their point of intersection is called the point of concurrence.
Equations of the given lines are
3x − 4y + 5 = 0 ...(i)
7x − 8y + 5 = 0 ...(ii)
4x + 5y − 45 = 0 ...(iii)
By (i) x 2 − (ii), we get
− x + 5 = 0
∴ x = 5
Substituting x = 5 in (i), we get
3(5) − 4y + 5 = 0
∴ −4y = − 20
∴ y = 5
∴ The point of intersection of lines (i) and (ii) is given by (5, 5).
Substituting x = 5 and y = 5 in L.H.S. of (iii), we get
L.H.S. = 4(5) + 5(5) − 45
= 20 + 25 − 45
= 0
= R.H.S.
∴ Line (iii) also passes through (5, 5).
Hence, the given three lines are concurrent and the point of concurrence is (5, 5).
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