Advertisements
Advertisements
प्रश्न
Find the equation of a line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.
Advertisements
उत्तर
Given lines
4x + 7y = 3 ...(1)
2x – 3y = – 1 ...(2)
(1) × 1 ⇒ 4x + 7y = 3 ...(3)
(2) × 2 ⇒ 4x – 6y = – 2 ...(4)
(–) (+) (+)
(3) – (4) ⇒ 13y = 5
y = `5/13`
Substitute the value of y = `5/13` in (2)
`2x - 3 xx 5/13` = – 1
`2x - 15/13` = – 1
26x – 15 = – 13
26x = – 13 + 15
26x = 2
x = `2/26 = 1/13`
The point of intersection is `(1/13, 5/13)`
Let the x-intercept and y-intercept be “a”
Equation of a line is
`x/"a" + y/"b"` = 1
`x/"a" + y/"b"` = 1 ...(equal intercepts)
It passes through `(1/13, 5/13)`
`1/(13"a") + 5/(13"a")` = 1
`(1 + 5)/(13"a")` = 1
13a = 6
a = `6/13`
The equation of the line is
`x/(6/13) + y/(6/13)` = 1
`(13x)/6 + (13y)/6` = 1
13x + 13y = 6
13x + 13y − 6 = 0
APPEARS IN
संबंधित प्रश्न
What is the inclination of a line whose slope is 1
What is the slope of a line perpendicular to the line joining A(5, 1) and P where P is the mid-point of the segment joining (4, 2) and (–6, 4).
Show that the given points are collinear: (– 3, – 4), (7, 2) and (12, 5)
Show that the given points form a right angled triangle and check whether they satisfy Pythagoras theorem
A(1, – 4), B(2, – 3) and C(4, – 7)
Show that the given points form a parallelogram:
A(2.5, 3.5), B(10, – 4), C(2.5, – 2.5) and D(– 5, 5)
Let A(3, – 4), B(9, – 4), C(5, – 7) and D(7, – 7). Show that ABCD is a trapezium.
A quadrilateral has vertices at A(– 4, – 2), B(5, – 1), C(6, 5) and D(– 7, 6). Show that the mid-points of its sides form a parallelogram.
The slope of the line joining (12, 3), (4, a) is `1/8`. The value of ‘a’ is
If slope of the line PQ is `1/sqrt(3)` then slope of the perpendicular bisector of PQ is
Without using distance formula, show that the points (−2, −1), (4, 0), (3, 3) and (−3, 2) are vertices of a parallelogram
