A-line-is-formed-by-joining-the-points-A-7-0-and-B-0-2-Obtain-the-equation-of-the-straight-line-joining-AC-such-that-the-x-axis-bisects-the-angle-BAC-

Question Number 170435 by MathsFan last updated on 23/May/22

$$\boldsymbol{{A}}\:\boldsymbol{{line}}\:\boldsymbol{{is}}\:\boldsymbol{{formed}}\:\boldsymbol{{by}}\:\boldsymbol{{joining}}\:\boldsymbol{{the}} \\ $$$$\boldsymbol{{points}}\:\boldsymbol{{A}}\left(\mathrm{7},\mathrm{0}\right)\:\boldsymbol{{and}}\:\boldsymbol{{B}}\left(\mathrm{0},\mathrm{2}\right).\:\boldsymbol{{Obtain}} \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{equation}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{straight}}\:\boldsymbol{{line}} \\ $$$$\boldsymbol{{joining}}\:\boldsymbol{{AC}}\:\boldsymbol{{such}}\:\boldsymbol{{that}}\:\boldsymbol{{the}}\:\boldsymbol{{x}}−\boldsymbol{{axis}} \\ $$$$\boldsymbol{{bisects}}\:\boldsymbol{{the}}\:\boldsymbol{{angle}}\:\boldsymbol{{BAC}}. \\ $$

Answered by MikeH last updated on 23/May/22

If the x−axis bisects ∠BAC ⇒ C(0,−2) m = ((△y)/(△x)) = ((−2−0)/(0−7)) = (2/7) y−0 = (2/7)(x−7) ⇒ 7y = 2x−14 (equation of line AC)

$$\mathrm{If}\:\mathrm{the}\:\mathrm{x}−\mathrm{axis}\:\mathrm{bisects}\:\angle{BAC} \\ $$$$\Rightarrow\:{C}\left(\mathrm{0},−\mathrm{2}\right) \\ $$$${m}\:=\:\frac{\bigtriangleup{y}}{\bigtriangleup{x}}\:=\:\frac{−\mathrm{2}−\mathrm{0}}{\mathrm{0}−\mathrm{7}}\:=\:\frac{\mathrm{2}}{\mathrm{7}} \\ $$$${y}−\mathrm{0}\:=\:\frac{\mathrm{2}}{\mathrm{7}}\left({x}−\mathrm{7}\right) \\ $$$$\Rightarrow\:\mathrm{7}{y}\:=\:\mathrm{2}{x}−\mathrm{14}\:\left(\mathrm{equation}\:\mathrm{of}\:\mathrm{line}\:{AC}\right) \\ $$

Commented by MathsFan last updated on 23/May/22

$${thanks}\:{but}\:{how}\:{did}\:{you}\:{get}\:{C}\left(\mathrm{0},−\mathrm{2}\right) \\ $$

Commented by MikeH last updated on 24/May/22

Commented by MikeH last updated on 24/May/22

looking from the sketch above, you find the angle BAC bisected by the x−axis but for the bisection(symmetric division) to occur, the point C has coordinates (0,−2).

$$\mathrm{looking}\:\mathrm{from}\:\mathrm{the}\:\mathrm{sketch}\:\mathrm{above},\:\mathrm{you}\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{angle}\:{BAC}\:\mathrm{bisected}\:\mathrm{by}\:\mathrm{the}\:\mathrm{x}−\mathrm{axis} \\ $$$$\mathrm{but}\:\mathrm{for}\:\mathrm{the}\:\mathrm{bisection}\left(\mathrm{symmetric}\:\mathrm{division}\right) \\ $$$$\mathrm{to}\:\mathrm{occur},\:\mathrm{the}\:\mathrm{point}\:\mathrm{C}\:\mathrm{has}\:\mathrm{coordinates} \\ $$$$\left(\mathrm{0},−\mathrm{2}\right).\: \\ $$

Commented by MathsFan last updated on 24/May/22

$${thank}\:{you}\:{very}\:{much}\:{sir}. \\ $$

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