Q.The distance of the point with position vector 3𝑖̂ + 4𝑗̂ + 5𝑘̂ from the y-axis is
(A) 4 units
(B) √34 units
(C) 5 units
(D) 5√2 units

Concept and Intuition

The foot of the perpendicular from any point $(x,y)$ onto the y-axis is simply $(0,y)$; so the distance asked for is just $|y|$. We only need P's $y$-coordinate, found by moving a known distance along a line of known slope from a known point.

Step-by-Step Solution

1. Solve the two line equations together: adding $\sqrt3x-y+2=0$ and $\sqrt3x+y-2=0$ gives $2\sqrt3x=0\Rightarrow x=0$, then $y=2$. So the intersection is $(0,2)$.
2. Line $\sqrt3x-y+2=0$ can be written $y=\sqrt3x+2$, slope $\sqrt3$, so its unit direction vector is $\left(\dfrac12,\dfrac{\sqrt3}2\right)$ (since $\sqrt{1^2+(\sqrt3)^2}=2$).
3. A point at distance 5 from $(0,2)$ along this line is $(0,2)+5\left(\dfrac12,\dfrac{\sqrt3}2\right)=\left(\dfrac52,\,2+\dfrac{5\sqrt3}2\right)$ (taking the positive direction).
4. The foot of the perpendicular from this point onto the y-axis is $\left(0,\,2+\dfrac{5\sqrt3}2\right)$. …

Concept and Intuition

A point equidistant from the x-axis and y-axis satisfies $|x|=|y|$, i.e. it lies on $y=x$ or $y=-x$. We intersect the given line with each of these two lines and see which quadrant each intersection falls in.

Step-by-Step Solution

1. Case $y=x$: substitute into $3x+5y=15$: $3x+5x=15 \Rightarrow 8x=15 \Rightarrow x=15/8$. So $y=15/8$, point $(15/8,15/8)$, both coordinates positive $\Rightarrow$ 1st quadrant.
2. Case $y=-x$: substitute: $3x-5x=15 \Rightarrow -2x=15 \Rightarrow x=-15/2$, so $y=15/2$, point $(-15/2,15/2)$: $x<0,y>0 \Rightarrow$ 2nd quadrant. …

Concept and Intuition

The distance of a point $(x,y,z)$ from the X-axis is $\sqrt{y^2+z^2}$ (drop the coordinate along that axis and use the other two); similarly for Y- and Z-axes. This problem is just careful bookkeeping of these squared distances.

Step-by-Step Solution

1. $d_1^2$ (distance$^2$ from X-axis) $=y^2+z^2=2^2+3^2=13$.
2. $d_2^2$ (from Y-axis) $=x^2+z^2=1^2+3^2=10$.
3. $d_3^2$ (from Z-axis) $=x^2+y^2=1^2+2^2=5$.
4. $2d_2^2+d_3^2+1=2(10)+5+1=26$. …

Concept and Intuition

Distance from a point $(x,y)$ to the $x$-axis is $|y|$ and to the $y$-axis is $|x|$. Equidistant means $|x|=|y|$, which splits into the two lines $y=x$ and $y=-x$.

Step-by-Step Solution

1. Case $y=x$: substitute into $3x+5y=15$: $3x+5x=15 \Rightarrow 8x=15 \Rightarrow x=\dfrac{15}{8}$, so $y=\dfrac{15}{8}$. Both coordinates positive → first quadrant.
2. Case $y=-x$: substitute: $3x-5x=15 \Rightarrow -2x=15 \Rightarrow x=-\dfrac{15}{2}$, so $y=\dfrac{15}{2}$. $x$ negative, $y$ positive → second quadrant. …

Concept and Intuition

The distance of a point $P(x,y)$ from the $Y$-axis is simply $|x|$. The condition $PA=2\times(\text{distance from }Y\text{-axis})$ becomes an equation relating $x,y$ once both distances are squared to remove the square roots and absolute value.

Step-by-Step Solution

1. Let $P=(x,y)$, $A=(2,-2)$. $PA=\sqrt{(x-2)^2+(y+2)^2}$; distance from $Y$-axis $=|x|$.
2. Condition: $\sqrt{(x-2)^2+(y+2)^2}=2|x|$.
3. Square both sides: $(x-2)^2+(y+2)^2=4x^2$.
4. Expand: $x^2-4x+4+y^2+4y+4=4x^2$. …

Concept and Intuition

A circle tangent to the Y-axis has its centre exactly one radius away from that axis, i.e. centre $=(\pm r, k)$ for centre height $k$. The length of the chord a circle cuts on a line at perpendicular distance $d$ from the centre is $2\sqrt{r^2-d^2}$ — applying this to the X-axis (distance $=|k|$ from centre) gives an equation for $r$.

Step-by-Step Solution

1. Circle touches Y-axis at $(0,3)$ $\Rightarrow$ centre $C=(h,3)$ with $|h|=r$ (radius).
2. X-axis intercept length 8: distance from centre to X-axis is $3$, so $2\sqrt{r^2-3^2}=8\Rightarrow\sqrt{r^2-9}=4\Rightarrow r^2=25\Rightarrow r=5$.
3. Since $|h|=r=5$ and $C$ is in the second quadrant ($h<0$, $k>0$): $C=(-5,3)$. …

Concept and Intuition

"Equidistant from the coordinate axes" means $|x|=|y|$; combined with being in the fourth quadrant ($x>0,\,y<0$), this forces the specific relation $x=-y$ (not $x=y$, which would put the point in quadrants I or III). Substituting the solved intersection point into $x=-y$ gives the required algebraic condition on $a,b,c,d$.

Step-by-Step Solution

1. Lines: $4ax+2ay+c=0$ and $5bx+2by+d=0$.
2. Solve simultaneously (Cramer's rule with determinant $D=4a\cdot2b-2a\cdot5b=-2ab$): $x=\dfrac{bc-ad}{ab}$, $y=\dfrac{4ad-5bc}{2ab}$.
3. Fourth quadrant means $x>0,y<0$; equidistant from axes means $|x|=|y|$, and together these force $x=-y$. …

Concept and Intuition

A line's $x$-intercept is found by setting $y=0$ and solving for $x$; the $y$-intercept by setting $x=0$ and solving for $y$. These are direct substitutions into the line's equation.

Step-by-Step Solution

1. Line: $5x-2y=10$.
2. $x$-intercept: set $y=0$: $5x=10 \Rightarrow x=2$.
3. $y$-intercept: set $x=0$: $-2y=10 \Rightarrow y=-5$.
4. Sum of squares of intercepts: $2^2+(-5)^2 = 4+25=29$.

Common Mistakes …

Concept and Intuition

A pair of lines (or any conic) meets the $y$-axis where $x=0$; this gives a quadratic in $y$ whose two roots are the $y$-coordinates of the two intersection points. The distance between them is the intercept length.

Step-by-Step Solution

1. Set $x=0$: $2(0)+4(0)y-4y^2-6(0)-8y+7=0 \Rightarrow -4y^2-8y+7=0$.
2. Multiply by $-1$: $4y^2+8y-7=0$.
3. Roots: $y=\dfrac{-8\pm\sqrt{64+112}}{8}=\dfrac{-8\pm\sqrt{176}}{8}=\dfrac{-2\pm\sqrt{11}}{2}$ (using $\sqrt{176}=4\sqrt{11}$). …