Concept understanding — Straight Lines — Forms of the Equation
The general (linear) equation of a straight line is ax+by+c=0, where a,b are not both zero; the set of solutions of any such equation is a straight line in the plane. Because dividing through by b (or a) removes one constant, every line's equation genuinely contains only two independent arbitrary constants — so exactly two independent pieces of information (two points, or a point and a slope, or two intercepts, etc.) are enough to pin a line down uniquely.
Slope. The angle of inclinationθ of a line is the angle it makes with the x-axis, measured counter-clockwise; the slopem=tanθ (undefined when θ=π/2, i.e. for a vertical line). Equivalently, through two points (x1,y1),(x2,y2) with x1=x2, m=x2−x1y2−y1; from the general form, m=−a/b (b=0). Three points are collinear exactly when the slope of any one pair equals the slope of another pair sharing a point.
Intercepts. The x-intercept is where a line meets the x-axis (y=0); the y-intercept is where it meets the y-axis (x=0). (x=0 is itself the equation of the y-axis; y=0 is the equation of the x-axis.)
The six forms (two conditions each, all interconvertible by algebra):
Let the x-intercept be a and y-intercept 1−a (sum =1), substitute the point (8,3) into the intercept form, and solve the resulting quadratic — both roots give valid lines.
Since the intercepts sum to 1, write them as a (on the x-axis) and 1−a (on the y-axis) for an unknown a.
Step 1. Write the intercept-form equation.
ax+1−ay=1
Step 2. Force the line through (8,3).
a8+1−a3=1
Step 3. Clear denominators (multiply by a(1−a)).
8(1−a)+3a=a(1−a)
8−8a+3a=a−a2
8−5a=a−a2
a2−6a+8=0
Step 4. Solve the quadratic.
(a−2)(a−4)=0⇒a=2ora=4
Both values are valid (neither makes an intercept zero), so there are two lines satisfying the condition.
Assuming there is only one line because the problem gives just one constraint value — the sum-of-intercepts condition is quadratic in the intercept, so two lines generally satisfy it. …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2026Set ANNUAL1 markMCQ
Q.The image of the point (2,3) in the line y=−x is:
(a) (−2,−3)
(b) (−3,−2)
(c) (3,2)
(d) (−3,2)
›Reveal solutionSolution
Reflecting (x,y) in the line y=−x gives (−y,−x), so (2,3)→(−3,−2).
The line y=−x passes through the origin at 135∘ to the x-axis. Reflection in this line swaps the coordinates and negates both: the general rule is (x,y)↦(−y,−x).