Question : Find the coordinates of the points where the graph $57x – 19y = 399$ cuts the coordinate axes.
Option 1: x-axis at(–7, 0) and y-axis at (0, –21)
Option 2: x-axis at(–7, 0) and y-axis at (0, 21)
Option 3: x-axis at (7, 0) and y-axis at (0, –21)
Option 4: x-axis at (7, 0) and y-axis at (0, 21)
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Correct Answer: x-axis at (7, 0) and y-axis at (0, –21)
Solution : $57x-19y=399$ ⇒ $\frac{57x}{399}-\frac{19y}{399} = 1$ ⇒ $\frac{x}{7}-\frac{y}{21} = 1$ -------------(i) Comparing it with the equation of a line: $\frac{x}{a}+\frac{y}{b} = 1$ x-intercept = $a$ = 7 y-intercept = $b$ = –21 The line, $57x-19y=399$, cuts the x-axis at (7, 0) and the y-axis at (0, –21) Hence, the correct answer is 'x-axis at (7, 0) and y-axis at (0, –21)'.
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Question : At what point does the line $2x–3y = 6$ cuts the Y-axis?
Option 1: (0, 2)
Option 2: (–2, 0)
Option 3: (2, 0)
Option 4: (0, –2)
Question : What is the equation of the line whose y-intercept is 4 and is making an angle of 45° with the positive x-axis?
Option 1: $x+y=–4$
Option 2: $x–y=4$
Option 3: $x+y=4$
Option 4: $x–y= – 4$
Question : The graph of the equation $x=a(a \neq 0)$ is a_____.
Option 1: line parallel to the x-axis
Option 2: line parallel to the y-axis
Option 3: line at an angle of 45 degree to y-axis
Option 4: line at an angle of 45 degree to x-axis
Question : The graphs of the equations $4 x+\frac{1}{3} y=\frac{8}{3}$ and $\frac{1}{2} x+\frac{3}{4} y+\frac{5}{2}=0$ intersect at a point P. The point P also lies on the graph of the equation:
Option 1: $x + 2y - 5 = 0$
Option 2: $3x - y - 7 = 0$
Option 3: $x - 3y - 12= 0$
Option 4: $4x - y + 7= 0$
Question : If $x^2+y^2=29$ and $xy=10$, where $x>0,y>0$ and $x>y$. Then the value of $\frac{x+y}{x-y}$ is:
Option 1: $- \frac{7}{3}$
Option 2: $\frac{7}{3}$
Option 3: $\frac{3}{7}$
Option 4: $-\frac{3}{7}$
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